Circle geometry

1. Parts of circles


ID is: 1567 Seed is: 6399

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Tangent
Chord (non-diameter)
Arc
2 attempts remaining
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Tangent F
Chord (non-diameter) B
Arc A
NOTE: a diameter is also a chord. However, this question asks you to identify a chord which is not a diameter, so the correct answer for Chord must be "B" and cannot be "D."
STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

ID is: 1567 Seed is: 612

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Diameter
Radius
Segment
2 attempts remaining
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Diameter D
Radius C
Segment E

STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

ID is: 1567 Seed is: 8366

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Arc
Radius
Tangent
2 attempts remaining
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Arc A
Radius C
Tangent F

STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

ID is: 1566 Seed is: 3183

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
A line that makes contact with a circle at one point on the circle.
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle.
The part of the circle that is cut off by a chord.
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
A line that makes contact with a circle at one point on the circle. Tangent
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle. Diameter
The part of the circle that is cut off by a chord. Segment

Submit your answer as: andand

ID is: 1566 Seed is: 2561

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
A part of the circumference of a circle.
The part of the circle that is cut off by a chord.
Any straight line segment from the centre of the circle to a point on the circumference.
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
A part of the circumference of a circle. Arc
The part of the circle that is cut off by a chord. Segment
Any straight line segment from the centre of the circle to a point on the circumference. Radius

Submit your answer as: andand

ID is: 1566 Seed is: 2440

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
A straight line segment joining the ends of an arc.
A line that makes contact with a circle at one point on the circle.
A part of the circumference of a circle.
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
A straight line segment joining the ends of an arc. Chord
A line that makes contact with a circle at one point on the circle. Tangent
A part of the circumference of a circle. Arc

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ID is: 1570 Seed is: 3251

Vocabulary: subtends

The figure below shows a circle. There are two points labelled on the circumference, A, and B. There are also two angles, labelled z and x.

Chord AB subtends which angle(s)?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to chord AB.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about chord AB: this chord is highlighted below in dark red. It stretches from point A to point B.

There are two shaded sections: one in blue and one with bars which are light blue. The chord subtends both of these angles - both of the angles spread out to meet the chord.

Therefore the correct answer is that chord AB subtends both of the angles in the figure: angle x and angle z.


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ID is: 1570 Seed is: 2952

Vocabulary: subtends

The figure below shows a circle. There are two points labelled on the circumference, W, and X. There are also two angles, labelled c and a.

Arc WX subtends which angle(s)?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to arc WX.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about arc WX: this arc is highlighted below in dark red. It stretches from point W to point X.

There are two shaded sections: one in blue and one with bars which are light blue. The arc subtends both of these angles - both of the angles spread out to meet the arc.

Therefore the correct answer is that arc WX subtends both of the angles in the figure: angle a and angle c.


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ID is: 1570 Seed is: 1535

Vocabulary: subtends

The figure below shows a circle. There are three points labelled on the circumference, G,H, and J. There are also two angles, labelled e and n.

Angle e is subtended by which arc?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' Does angle e spread out to meet arc GH or arc HJ, or some other arc?


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us to find the arc which is connected to e.

The shaded part of the figure (in blue) shows that angle e spreads out to meet arc GH. (This arc is shown in dark red from point G to point H.)

Therefore the correct answer is that angle e is subtended by arc GH.


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2. Chord theorems


ID is: 3933 Seed is: 9952

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. S and T are points on AC and AB respectively such that OSAC and OTAB. AB=40 and AC=48.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AS.

    Answer:

    AS=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AS. We know that AC has a length of 48. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OS is perpendicular to chord AC. Therefore OS bisects AC. And that means AS is half as long as AC. We can use that fact to find the length of AS.

    AS=AC2=482=24

    AS has a length of 24.


    Submit your answer as:
  2. If OS=715OT, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOTA, side TA must be half as long as AB. This is for the same reason we used in Question 1. So TA=20, as labelled below. We are using d for the length of OT.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOSA:

    c2=a2+b2r2=OS2+SA2r2=(715d)2+(24)2

    We can write a similar equation for ΔOTA.

    c2=a2+b2r2=OT2+TA2r2=(d)2+(20)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+202=(715d)2+242
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OT).

    d2+202=(715)2d2+242d2+400=49225d2+576d249225d2=576400176225d2=176d2=225d=±15

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(20)2r2=(15)2+400r2=225+400=625r=±25

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 25.


    Submit your answer as:

ID is: 3933 Seed is: 6358

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. S and T are points on AC and AB respectively such that OSAC and OTAB. AB=136 and AC=154.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AS.

    Answer:

    AS=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AS. We know that AC has a length of 154. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OS is perpendicular to chord AC. Therefore OS bisects AC. And that means AS is half as long as AC. We can use that fact to find the length of AS.

    AS=AC2=1542=77

    AS has a length of 77.


    Submit your answer as:
  2. If OS=1217OT, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOTA, side TA must be half as long as AB. This is for the same reason we used in Question 1. So TA=68, as labelled below. We are using d for the length of OT.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOSA:

    c2=a2+b2r2=OS2+SA2r2=(1217d)2+(77)2

    We can write a similar equation for ΔOTA.

    c2=a2+b2r2=OT2+TA2r2=(d)2+(68)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+682=(1217d)2+772
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OT).

    d2+682=(1217)2d2+772d2+4624=144289d2+5929d2144289d2=59294624145289d2=1305d2=2601d=±51

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(68)2r2=(51)2+4624r2=2601+4624=7225r=±85

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 85.


    Submit your answer as:

ID is: 3933 Seed is: 3131

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. X and Y are points on AC and AB respectively such that OXAC and OYAB. AB=40 and AC=48.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AY.

    Answer:

    AY=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AY. We know that AB has a length of 40. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OY is perpendicular to chord AB. Therefore OY bisects AB. And that means AY is half as long as AB. We can use that fact to find the length of AY.

    AY=AB2=402=20

    AY has a length of 20.


    Submit your answer as:
  2. If OX=715OY, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOXA, side XA must be half as long as AC. This is for the same reason we used in Question 1. So XA=24, as labelled below. We are using d for the length of OY.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOXA:

    c2=a2+b2r2=OX2+XA2r2=(715d)2+(24)2

    We can write a similar equation for ΔOYA.

    c2=a2+b2r2=OY2+YA2r2=(d)2+(20)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+202=(715d)2+242
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OY).

    d2+202=(715)2d2+242d2+400=49225d2+576d249225d2=576400176225d2=176d2=225d=±15

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(20)2r2=(15)2+400r2=225+400=625r=±25

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 25.


    Submit your answer as:

ID is: 1626 Seed is: 983

Using the perpendicular bisector theorem

In the circle with centre O, segment JK¯=24. OE¯ is perpendicular to JK¯ at Point E. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    EK¯ is because .

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, JK¯ is the chord, and it is bisected at point E. In other words, E is the midpoint of JK¯. We know the chord is bisected because it is perpendicular to segment OE¯.

    On the diagram below the segments JE¯ and EK¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: EK¯ is "half the length of JK¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of EK¯ in the diagram.

    Answer: EK¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is smaller than the length given. Therefore, we must divide by two to get the length of the smaller segment (this is the same as multiplying by a half).

    smaller=12×biggerEK¯=12(JK¯)=12(24)=12

    The length of EK¯ is 12.


    Submit your answer as:

ID is: 1626 Seed is: 1826

Using the perpendicular bisector theorem

In the circle with centre O, segment FJ¯=6. OF¯ is perpendicular to KJ¯ at Point F. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    KJ¯ is because .

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, KJ¯ is the chord, and it is bisected at point F. In other words, F is the midpoint of KJ¯. We know the chord is bisected because it is perpendicular to segment OF¯.

    On the diagram below the segments KF¯ and FJ¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: KJ¯ is "twice as long as FJ¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of KJ¯ in the diagram.

    Answer: KJ¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is larger than the length given. Therefore, we must multiply by two to get the length of the chord.

    larger=2×shorterKJ¯=2(FJ¯)=2(6)=12

    The length of KJ¯ is 12.


    Submit your answer as:

ID is: 1626 Seed is: 4981

Using the perpendicular bisector theorem

In the circle with centre O, segment KA¯=15. OA¯ is perpendicular to KJ¯ at Point A. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    KJ¯ is because .

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, KJ¯ is the chord, and it is bisected at point A. In other words, A is the midpoint of KJ¯. We know the chord is bisected because it is perpendicular to segment OA¯.

    On the diagram below the segments KA¯ and AJ¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: KJ¯ is "twice as long as KA¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of KJ¯ in the diagram.

    Answer: KJ¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is larger than the length given. Therefore, we must multiply by two to get the length of the chord.

    larger=2×smallerKJ¯=2(KA¯)=2(15)=30

    The length of KJ¯ is 30.


    Submit your answer as:

ID is: 1625 Seed is: 5877

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord AP¯=8 and OP¯=6. Also, OB¯AP¯ at point B. Compute the length of OB¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OB¯ is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord AP¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of AP¯=8
  • length of OP¯=6 which is the radius of the circle
  • OB¯ is perpendicular to AP¯

and we want:

  • the length of the side OB¯

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord AP¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point B. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that AB¯ and BP¯ are half the size of AP¯, which is 8.

AB¯=BP¯=AP¯2(line from centre bisects chord)=(8)2=4

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OP¯ is the radius .

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OB¯, the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2OP¯2=OB¯2+BP¯2(6)2=OB¯2+(4)2OB¯=±3616=±20

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OB¯=4.47.


Submit your answer as:

ID is: 1625 Seed is: 7217

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord KR¯=7 and OC¯=5.5. Also, OC¯KR¯ at point G. Compute the length of OG¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OG¯ is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord KR¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of KR¯=7
  • length of OC¯=5.5 which is the radius of the circle
  • OC¯ is perpendicular to KR¯

and we want:

  • the length of the side OG¯

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord KR¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point G. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that KG¯ and GR¯ are half the size of KR¯, which is 7.

KG¯=GR¯=KR¯2(line from centre bisects chord)=(7)2=3.5

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OC¯ is the radius of the circle; and more importantly, it is equal to any other radius of the circle, e.g. OC¯OK¯.

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OG¯, the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2OK¯2=OG¯2+GK¯2(5.5)2=OG¯2+(3.5)2OG¯=±(30.25)(12.25)=±18

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OG¯=4.24.


Submit your answer as:

ID is: 1625 Seed is: 4590

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord DC¯=7 and OC¯=4.5. Also, OR¯DC¯ at point R. Compute the length of OR¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OR¯ is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord DC¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of DC¯=7
  • length of OC¯=4.5 which is the radius of the circle
  • OR¯ is perpendicular to DC¯

and we want:

  • the length of the side OR¯

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord DC¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point R. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that DR¯ and RC¯ are half the size of DC¯, which is 7.

DR¯=RC¯=DC¯2(line from centre bisects chord)=(7)2=3.5

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OC¯ is the radius .

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OR¯, the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2OC¯2=OR¯2+RC¯2(4.5)2=OR¯2+(3.5)2OR¯=±20.2512.25=±8

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OR¯=2.83.


Submit your answer as:

ID is: 1627 Seed is: 9249

Using circle theorems to relate angles

In the circle with midpoint M the arc QA subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is QA and the angles x and y are both subtended by it. Therefore, the angles must be the same size: angle y in terms of x is y=x.

The reason for this theorem is: s subtended by same arc QA.


Submit your answer as: and

ID is: 1627 Seed is: 2819

Using circle theorems to relate angles

In the circle with midpoint M the arc CJ subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

A useful circle theorem for this question is the following: the angle subtended by an arc at the centre of a circle is twice the size of the angle subtended by the same arc at the circumference of the circle. Here the arc is CJ; the angle subtended by the arc at the centre of a circle is x; and the angle subtended by the same arc at the circumference of the circle is y. Therefore, x is twice the size of y or: x=2y. Writing angle y in terms of x, we get y=12x.

The reason for this theorem is:  at centre =2 at circum.


Submit your answer as: and

ID is: 1627 Seed is: 727

Using circle theorems to relate angles

In the circle with midpoint M the arc KB subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is KB and the angles x and y are both subtended by it. Therefore, the angles must be the same size: angle y in terms of x is y=x.

The reason for this theorem is: s subtended by same arc KB.


Submit your answer as: and

ID is: 4364 Seed is: 1064

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, O is the centre of the circle. J, K and L are points on the circumference of the circle. KO is produced to M. L^=y and J^=x.

  1. Write an expression for O^1 in terms of y.

    Answer:

    OL and OK are radii.
    =y (s opp equal sides)
    O1=

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Radii are always the same length. ΔLOK has two sides that are radii.


    STEP: Two sides in ΔLOK are radii
    [−1 point ⇒ 2 / 3 points left]
    OL=OK (radii)K^1=y(s opp equal sides)

    STEP: O^1 is an exterior angle of ΔLOK
    [−2 points ⇒ 0 / 3 points left]
    The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext of Δ).
    O^1=y+y(ext  of Δ)=2y

    Submit your answer as: andand
  2. Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states reflex LO^J=2LK^J.

    Here's the diagram again, for your convenience:

    Answer:

    In the same way as in Question 1, O2=2x.

    reflex LOJ
    =
    =
    =
    =2LKJ

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    reflex LO^J=O^1+O^2

    Use the expressions that you have for O^1 and O^2.


    STEP: Use the expressions for O^1 and O^2, then take out the common factor
    [−2 points ⇒ 1 / 3 points left]

    We are trying to prove that reflex LO^J is double the size of LK^J. We are going to split up both angles into two parts and work with the sum.

    In the same way as in Question 1, O^2=2x.

    reflex LO^J=2y+2x=2(y+x)

    STEP: Replace y with K^1 and x with K^2, since they are equal
    [−1 point ⇒ 0 / 3 points left]
    reflex LO^J=2(K^1+K^2)=2LK^J

    Submit your answer as: andand

ID is: 4364 Seed is: 4561

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, O is the centre of the circle. J, K and L are points on the circumference of the circle. KO is produced to M. L^=y and J^=x.

  1. Write an expression for O^2 in terms of x.

    Answer:

    OJ and OK are radii.
    =x (s opp equal sides)
    O2=

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Radii are always the same length. ΔJOK has two sides that are radii.


    STEP: Two sides in ΔJOK are radii
    [−1 point ⇒ 2 / 3 points left]
    OJ=OK (radii)K^2=x(s opp equal sides)

    STEP: O^2 is an exterior angle of ΔJOK
    [−2 points ⇒ 0 / 3 points left]
    The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext of Δ).
    O^2=x+x(ext  of Δ)=2x

    Submit your answer as: andand
  2. Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states LO^J=2LK^J.

    Here's the diagram again, for your convenience:

    Answer:

    In the same way as in Question 1, O1=2y.

    LOJ
    =
    =
    =
    =2LKJ

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
     LO^J=O^1+O^2

    Use the expressions that you have for O^1 and O^2.


    STEP: Use the expressions for O^1 and O^2, then take out the common factor
    [−2 points ⇒ 1 / 3 points left]

    We are trying to prove that LO^J is double the size of LK^J. We are going to split up both angles into two parts and work with the sum.

    In the same way as in Question 1, O^1=2y.

    LO^J=2x+2y=2(x+y)

    STEP: Replace x with K^2 and y with K^1, since they are equal
    [−1 point ⇒ 0 / 3 points left]
    LO^J=2(K^2+K^1)=2LK^J

    Submit your answer as: andand

ID is: 4364 Seed is: 2957

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, O is the centre of the circle. A, B and C are points on the circumference of the circle. CO is produced to D. B^=y and A^=x.

  1. Write an expression for O^1 in terms of y.

    Answer:

    OB and OC are radii.
    C^1=y
    O^1=

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Radii are always the same length. ΔBOC has two sides that are radii.


    STEP: Two sides in ΔBOC are radii
    [−1 point ⇒ 2 / 3 points left]
    OB=OC (radii)C^1=y(s opp equal sides)

    STEP: O^1 is an exterior angle of ΔBOC
    [−2 points ⇒ 0 / 3 points left]
    The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext of Δ).
    O^1=y+y(ext  of Δ)=2y

    Submit your answer as: andand
  2. Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states reflex BO^A=2BC^A.

    Here's the diagram again, for your convenience:

    Answer:

    In the same way as in Question 1, O2=2x.

    reflex BOA
    =
    =
    =
    =2BCA

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    reflex BO^A=O^1+O^2

    Use the expressions that you have for O^1 and O^2.


    STEP: Use the expressions for O^1 and O^2, then take out the common factor
    [−2 points ⇒ 1 / 3 points left]

    We are trying to prove that reflex BO^A is double the size of BC^A. We are going to split up both angles into two parts and work with the sum.

    In the same way as in Question 1, O^2=2x.

    reflex BO^A=2x+2y=2(x+y)

    STEP: Replace x with C^2 and y with C^1, since they are equal
    [−1 point ⇒ 0 / 3 points left]
    reflex BO^A=2(C^2+C^1)=2BC^A

    Submit your answer as: andand

ID is: 4386 Seed is: 831

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, O is the centre of the circle. Diameter BD subtends BA^D at the circumference of the circle. C is another point on the circumference and chords BC and AC are drawn such that AC intersects BD at O. C^=31°.

Calculate, giving reasons, the size of:

Answer:
  1. BA^D= °
  2. D^= °
  3. BO^A= °
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!


STEP: Find BA^D using diameter BD
[−2 points ⇒ 4 / 6 points left]

We were given that BD is a diameter. It subtends the angle BA^D at the circumference.

BA^D=90° ( in semi circle)

Choose the right reason!

  • When we know that a chord is a diameter, we use in semi circle to prove that an angle is 90°.
  • When we know that an angle is 90°, we use chord subtends 90° to prove that the chord is a diameter.

STEP: Find D^ using angles in the same segment
[−2 points ⇒ 2 / 6 points left]

The angles C^ and D^ are both subtended by the chord BA at the circumference.

D^=C^=31° (s in the same seg)


STEP: Find BO^A using angle at centre = 2× angle at circumference
[−2 points ⇒ 0 / 6 points left]

The angles BO^A and C^ are both subtended by chord BA. BO^A is at the centre and C^ is at the circumference.

2×31°=BO^A ( at centre =2× at circumference)

BO^A=62°


Submit your answer as: andandandandand

ID is: 4386 Seed is: 2517

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, O is the centre of the circle. Diameter FH subtends FE^H at the circumference of the circle. G is another point on the circumference and chords FG and EG are drawn such that EG intersects FH at O. G^=53°.

Calculate, giving reasons, the size of:

Answer:
  1. FE^H= °
  2. FO^E= °
  3. F^2= °
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!


STEP: Find FE^H using diameter FH
[−2 points ⇒ 4 / 6 points left]

We were given that FH is a diameter. It subtends the angle FE^H at the circumference.

FE^H=90° ( in semi circle)

Choose the right reason!

  • When we know that a chord is a diameter, we use in semi circle to prove that an angle is 90°.
  • When we know that an angle is 90°, we use chord subtends 90° to prove that the chord is a diameter.

STEP: Find FO^E using angle at centre = 2× angle at circumference
[−2 points ⇒ 2 / 6 points left]

The angles FO^E and G^ are both subtended by chord FE. FO^E is at the centre and G^ is at the circumference.

2×53°=FO^E ( at centre =2× at circumference)

FO^E=106°


STEP: Find F^2 using ΔFOG
[−2 points ⇒ 0 / 6 points left]

OF=OG (radii)

F^2=53° (s opp equal sides)

NOTE: As usual in Geometry, there is more than one correct way of finding F^2, but we can only show you one way here. If you justify your answer using correct reasons, you will still be marked correct.

Submit your answer as: andandandandand

ID is: 4386 Seed is: 3180

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, O is the centre of the circle. Diameter QS subtends QP^S at the circumference of the circle. R is another point on the circumference and chords QR and PR are drawn. Radius PO is drawn. QO^P=80°.

Calculate, giving reasons, the size of:

Answer:
  1. QP^S= °
  2. S^= °
  3. Q^1= °
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!


STEP: Find QP^S using diameter QS
[−2 points ⇒ 4 / 6 points left]

We were given that QS is a diameter. It subtends the angle QP^S at the circumference.

QP^S=90° ( in semi circle)

Choose the right reason!

  • When we know that a chord is a diameter, we use in semi circle to prove that an angle is 90°.
  • When we know that an angle is 90°, we use chord subtends 90° to prove that the chord is a diameter.

STEP: Find S^ using angle at centre = 2× angle at circumference
[−2 points ⇒ 2 / 6 points left]

The angles QO^P and S^ are both subtended by chord QP. QO^P is at the centre and S^ is at the circumference.

2×S^=80° ( at centre =2× at circumference)

S^=40°


STEP: Find Q^1 using ΔPQS
[−2 points ⇒ 0 / 6 points left]

Q^1=180°90°40° (sum of s in Δ)

Q^1=50°

NOTE: As usual in Geometry, there is more than one correct way of finding Q^1, but we can only show you one way here. If you justify your answer using correct reasons, you will still be marked correct.

Submit your answer as: andandandandand

3. Cyclic quadrilateral theorems


ID is: 1571 Seed is: 5097

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: A,B,C,D, and E. There is also one angle given: BC^D=135°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point A because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'EDCB' or 'CDEB' or 'BEDC' etc. However, you may not write 'edcb' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    BCDECDEBDEBCEBCDEDCBBEDCCBEDDCBE

    Submit your answer as:
  2. What are the measures of the angles DE^B and EB^C in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of DE^B= °
    2. The measure of EB^C=°
    numeric
    string
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: B^ and D^ are opposite each other; C^ and E^ are opposite from each other.

    We know that C^ and E^ have a sum of 180°:

    C^+E^=180°(135°)+E^=180°E^=180°135°E^=45°

    The same relationship is true for D^ and B^ (they are supplementary). However, we cannot find the answer for B^ because we don't know the value of D^. (All we can say is that together they make 180°.) So we cannot determine the measure for this angle.

    The answers are:

    1. The answer for DE^B is 45°.
    2. The answer for EB^C is no solution.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

ID is: 1571 Seed is: 5299

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: G,H,I,J, and K. There are also two angles given: J^=133° and K^=80°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point I because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'KGHJ' or 'GHJK' or 'GKJH' etc. However, you may not write 'kghj' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    GHJKHJKGJKGHKGHJKJHGGKJHHGKJJHGK

    Submit your answer as:
  2. What are the measures of the angles KG^H and GH^J in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of KG^H= °
    2. The measure of GH^J=°
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: G^ and J^ are opposite each other; H^ and K^ are opposite from each other.

    We know that J^ and G^ have a sum of 180°:

    J^+G^=180°(133°)+G^=180°G^=180°133°G^=47°

    It is the same story over again for angles K^ and H^H^=18080°=100°.

    The answers are:

    1. The answer for KG^H is 47°.
    2. The answer for GH^J is 100°.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

ID is: 1571 Seed is: 5800

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: G,H,I,J, and K. There are also two angles given: J^=45° and K^=120°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point G because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'HKJI' or 'JKHI' or 'HIJK' etc. However, you may not write 'hkji' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    HIJKIJKHJKHIKHIJKJIHHKJIIHKJJIHK

    Submit your answer as:
  2. What are the measures of the angles KH^I and HI^J in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of KH^I= °
    2. The measure of HI^J=°
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: H^ and J^ are opposite each other; I^ and K^ are opposite from each other.

    We know that J^ and H^ have a sum of 180°:

    J^+H^=180°(45°)+H^=180°H^=180°45°H^=135°

    It is the same story over again for angles K^ and I^I^=180120°=60°.

    The answers are:

    1. The answer for KH^I is 135°.
    2. The answer for HI^J is 60°.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

ID is: 4003 Seed is: 7673

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral WXYZ is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord at the centre of the circle is twice the angle subtended by the same chord the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that W+Y=180°.

    Answer:
    Statement Reason
    at centre = 2 at circum.
    O2=2Y at centre = 2 at circum.
    s in a rev.
    W+Y=180°
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2W at centre = 2 at circum.
    O2=2Y at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2W+2Y=360°
    W+Y=180°

    Submit your answer as: andand

ID is: 4003 Seed is: 3965

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral ABCD is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord at the centre of the circle is twice the angle subtended by the same chord the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that A+C=180°.

    Answer:
    Statement Reason
    O1=2A at centre = 2 at circum.
    at centre = 2 at circum.
    s in a rev.
    A+C=180°
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2A at centre = 2 at circum.
    O2=2C at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2A+2C=360°
    A+C=180°

    Submit your answer as: andand

ID is: 4003 Seed is: 9813

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral WXYZ is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord at the centre of the circle is twice the angle subtended by the same chord the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that W+Y=180°.

    Answer:
    Statement Reason
    O1=2W at centre = 2 at circum.
    at centre = 2 at circum.
    s in a rev.
    W+Y=180°
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2W at centre = 2 at circum.
    O2=2Y at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2W+2Y=360°
    W+Y=180°

    Submit your answer as: andand

ID is: 1565 Seed is: 1391

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 5 points: P,Q,R,S, and T.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point R because it is not on the circle.

    The figure shows 4 points on the circle, which are circled in red.

    Because the shape has four sides and the vertices are all on the edge of the circle, the figure PQST is a cyclic quadrilateral. The cyclic quadrilateral is shown shaded in light blue.

    The correct answer from the list is: PQST.


    Submit your answer as:
  2. In the cyclic quadrilateral, which angle is opposite to Q^?

    Answer: The angle is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Start at point Q and drag your finger across the quadrilateral to see which point is on the other side. When you do this, your finger will go between the other two points which make the quadrilateral.
    STEP: Identify pairs of opposite angles
    [−1 point ⇒ 0 / 1 points left]

    The quadrilateral has two pairs of opposite angles: P^ and S^ are opposite each other; Q^ and T^ are opposite from each other.

    The correct answer is T.


    Submit your answer as:

ID is: 1565 Seed is: 4933

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 5 points: U,V,W,X, and Y.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point U because it is not on the circle.

    The figure shows 4 points on the circle, which are circled in red.

    Because the shape has four sides and the vertices are all on the edge of the circle, the figure VWXY is a cyclic quadrilateral. The cyclic quadrilateral is shown shaded in light blue.

    The correct answer from the list is: VWXY.


    Submit your answer as:
  2. In the cyclic quadrilateral, which angle is opposite to V^?

    Answer: The angle is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Start at point V and drag your finger across the quadrilateral to see which point is on the other side. When you do this, your finger will go between the other two points which make the quadrilateral.
    STEP: Identify pairs of opposite angles
    [−1 point ⇒ 0 / 1 points left]

    The quadrilateral has two pairs of opposite angles: V^ and X^ are opposite each other; W^ and Y^ are opposite from each other.

    The correct answer is X.


    Submit your answer as:

ID is: 1565 Seed is: 2422

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 4 points: G,H,I, and J.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point G because it is not on the circle.

    The figure shows 2 points on the circle, which are circled in red. (Point H is inside of the circle, not on the circumference!)

    There is no cyclic shape in this figure.

    The correct answer from the list is: there is no cyclic quadrilateral.


    Submit your answer as:
  2. Which of the following statements is true about cyclic quadrilaterals?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    One important fact about cyclic quadrilaterals tells us about the size of opposite angles.
    STEP: Use the facts about angles in cyclic quadrilaterals
    [−1 point ⇒ 0 / 1 points left]

    You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

    The correct answer is opposite angles are supplementary.


    Submit your answer as:

ID is: 4371 Seed is: 4520

Overlapping circles: find equal angles

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, QN is a common chord of the two circles. The centre of the larger circle, M, lies on the circumference of the smaller circle. NP is produced to S, a point on the larger circle. PM produced meets the chord QS at R. Q^2=x.

  1. Give a reason why P^2=x.

    Answer:

    P^2=x because:

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What geometry reason can you use to link P^2 to x?


    STEP: Consider the relationship between Q^2 and P^2
    [−1 point ⇒ 0 / 1 points left]

    QMPN is a cyclic quadrilateral because all of its points lie on the circumference of the same circle. NS is a straight line, so P^2 is an exterior angle of the quadrilateral.

    P^2=x (ext  of cyclic quad)


    Submit your answer as:
  2. Determine another angle equal to x, giving reasons.

    Here's the diagram again:

    Answer:


    =x

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The radii of a circle are equal in length.


    STEP: Use s opposite equal sides in ΔMQN
    [−2 points ⇒ 0 / 2 points left]

    M is the centre of the larger circle, so MQ and MN are radii.

    MQ=MN(radii)
    N^1=x(s opp equal sides)
    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand

ID is: 4371 Seed is: 4467

Overlapping circles: find equal angles

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, BA is a common chord of the two circles. The centre of the larger circle, O, lies on the circumference of the smaller circle. AC is produced to E, a point on the larger circle. CO produced meets the chord BE at D. C^2=x.

  1. Give a reason why B^2=x.

    Answer:

    B^2=x because:

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What geometry reason can you use to link B^2 to x?


    STEP: Consider the relationship between C^2 and B^2
    [−1 point ⇒ 0 / 1 points left]

    BOCA is a cyclic quadrilateral because all of its points lie on the circumference of the same circle. AE is a straight line, so C^2 is an exterior angle of the quadrilateral.

    B^2=x (ext  of cyclic quad)


    Submit your answer as:
  2. Determine another angle equal to x, giving reasons.

    Here's the diagram again:

    Answer:


    =x

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The radii of a circle are equal in length.


    STEP: Use s opposite equal sides in ΔOBA
    [−2 points ⇒ 0 / 2 points left]

    O is the centre of the larger circle, so OB and OA are radii.

    OB=OA(radii)
    A^1=x(s opp equal sides)
    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand

ID is: 4371 Seed is: 5273

Overlapping circles: find equal angles

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, PN is a common chord of the two circles. The centre of the larger circle, M, lies on the circumference of the smaller circle. NQ is produced to S, a point on the larger circle. QM intersects PS at R. Q^2=x.

  1. Give a reason why NP^M=x.

    Answer:

    NP^M=x because:

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What geometry reason can you use to link NP^M to x?


    STEP: Consider the relationship between Q^2 and NP^M
    [−1 point ⇒ 0 / 1 points left]

    PMQN is a cyclic quadrilateral because all of its points lie on the circumference of the same circle. NS is a straight line, so Q^2 is an exterior angle of the quadrilateral.

    NP^M=x (ext  of cyclic quad)


    Submit your answer as:
  2. Determine another angle equal to x, giving reasons.

    Here's the diagram again:

    Answer:


    =x

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The radii of a circle are equal in length.


    STEP: Use s opposite equal sides in ΔMPN
    [−2 points ⇒ 0 / 2 points left]

    M is the centre of the larger circle, so MP and MN are radii.

    MP=MN(radii)
    N^1=x(s opp equal sides)
    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand

ID is: 4372 Seed is: 1598

Overlapping circles: algebraic questions

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, CF is a common chord of the two circles. FD is a diameter of the larger circle and O is the centre of the smaller circle. FD intersects AC at E and FB is a straight line.

C^2=A^=x.

  1. Determine an expression for C^1 in terms of x.

    Answer:

    BC^F=
    C^1= °

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use circle geometry theorems to determine an expression for C^1.


    STEP: Use in a semi-circle to determine BC^F
    [−2 points ⇒ 1 / 3 points left]

    BC^F=90° ( in semi circle)


    STEP: Subtract the angles to determine an expression for C^1
    [−1 point ⇒ 0 / 3 points left]

    From the diagram, we can see that

    BC^F=C^1+C^290°=C^1+xC^1=90°x

    Submit your answer as: andand
  2. Prove that AE=EC.

    Here's the diagram again:

    Answer:

    D^= (s in the same seg)
    E^2=90°
    AE=EC

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    There are two theorems that you usually use to prove that lines are equal in length:

    • If the angles in a triangle are equal in size, then the sides opposite those angles are equal in length (sides opp equal s).
    • If a line drawn from the centre of a circle is perpendicular to a chord, then it bisects the chord (line from centre to chord).

    STEP: Plan your approach
    [−1 point ⇒ 2 / 3 points left]

    There are two theorems that you usually use to prove that lines are equal in length:

    • If the angles in a triangle are equal in size, then the sides opposite those angles are equal in length (sides opp equal s).
    • If a line drawn from the centre of a circle is perpendicular to a chord, then it bisects the chord (line from centre to chord).

    We will prove that E^2=90°. Then we will use line from centre to chord to conclude that AE=EC.


    STEP: Determine D^ and hence prove that E^2=90°
    [−1 point ⇒ 1 / 3 points left]

    In Question 1, we proved that C^1=90°x.

    D^=90°x(s in the same seg)

    Now we can use sum of s in ΔEAD to determine E^2:

    E^2+x+(90°x)=180° (sum of s in Δ)

    E^2=90°

    STEP: Since DF is a diameter and DFAC, we can conclude that AE=EC
    [−1 point ⇒ 0 / 3 points left]

    AE=EC (line from centre  to chord)

    Choose the right reason! There are three similar theorems about a line from the centre to a chord:

    • To prove that a line from the centre bisects a chord, we must prove that the lines are perpendicular and use line from centre to chord.
    • To prove that a line from the centre is perpendicular to a chord, we must prove that the line is bisected and use line from centre to midpt of chord.
    • To prove that a line passes through the centre of a circle, we must prove that it is a perpendicular bisector of a chord and use perp bisector of chord.

    Submit your answer as: andand

ID is: 4372 Seed is: 3370

Overlapping circles: algebraic questions

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, the centre of the larger circle, M, lies on the circumference of the smaller circle. QN is produced to R, a point on the larger circle. QM is produced to L, a point on the larger circle.

L^=R^1=N^1=x.

  1. Determine an expression for Q^ in terms of x.

    Answer:

    M^3=180°2x
    Q^= °

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use circle geometry theorems to determine an expression for Q^.


    STEP: Use sum of s in ΔMLR to get an expression for M^3
    [−1 point ⇒ 2 / 3 points left]

    M^3+x+x=180° (sum of s in Δ)

    M^3=180°2x

    STEP: Use "angle at centre" to determine an expression for Q^
    [−2 points ⇒ 0 / 3 points left]

    Q^=180°2x2 ( at centre =2× at circumference)

    Q^=180°22x2=90°x
    TIP: If you were working on paper, you could also have answered this question using in semi circle in the big circle, then sum of s in ΔLQR. But, to answer this question you needed to complete the steps as they are written here.

    Submit your answer as: andand
  2. Prove that QN is a diameter of the smaller circle.

    Here's the diagram again:

    Answer:

    M^1=
    QN is a diameter

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    There are two theorems that you usually use to prove that a chord is a diameter:

    • If the angle subtended by a chord at the circumference of the circle is 90°, then the chord is a diameter (converse in semi circle) or (chord subtends 90°).
    • If a line is the perpendicular bisector of a chord, then it passes through the centre of the circle (perp bisector of chord). If that line is a chord, then it is a diameter.

    STEP: Plan your approach
    [−1 point ⇒ 2 / 3 points left]

    There are two theorems that you usually use to prove that a chord is a diameter:

    • If the angle subtended by a chord at the circumference of the circle is 90°, then the chord is a diameter (converse in semi circle) or (chord subtends 90°).
    • If a line is the perpendicular bisector of a chord, then it passes through the centre of the circle (perp bisector of chord). If that line is a chord, then it is a diameter.

    We will prove that M^1=90° and use converse in semi circle to conclude that QN is a diameter.


    STEP: Prove that M^1=90° using sum of s in ΔNMQ
    [−1 point ⇒ 1 / 3 points left]

    In Question 1, we proved that Q^=90°x.

    M^1+x+(90°x)=180° (sum of s in Δ)

    M^1=90°

    STEP: Now that we know that M^1=90°, we can conclude that QN is a diameter
    [−1 point ⇒ 0 / 3 points left]

    Therefore QN is a diameter of the smaller circle (converse in semi circle).

    TIP: You can also use chord subtends 90° as your reason. They are just two different abbreviations of the same theorem.

    Submit your answer as: andand

ID is: 4372 Seed is: 217

Overlapping circles: algebraic questions

Adapted from DBE Nov 2015 Grade 11, P2, Q10.2
Maths formulas

In the diagram, BD is a common chord of the two circles. DF is a diameter of the larger circle and O is the centre of the smaller circle. DF intersects CB at E and DA is a straight line.

B^2=C^=x.

  1. Determine an expression for B^1 in terms of x.

    Answer:

    =90°
    B1= °

    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use circle geometry theorems to determine an expression for B^1.


    STEP: Use in a semi-circle to determine AB^D
    [−2 points ⇒ 1 / 3 points left]

    AB^D=90° ( in semi circle)


    STEP: Subtract the angles to determine an expression for B^1
    [−1 point ⇒ 0 / 3 points left]

    From the diagram, we can see that

    AB^D=B^1+B^290°=B^1+xB^1=90°x

    Submit your answer as: andand
  2. Prove that CE=EB.

    Here's the diagram again:

    Answer:

    F^= (s in the same seg)
    E^2=90°
    CE=EB

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    There are two theorems that you usually use to prove that lines are equal in length:

    • If the angles in a triangle are equal in size, then the sides opposite those angles are equal in length (sides opp equal s).
    • If a line drawn from the centre of a circle is perpendicular to a chord, then it bisects the chord (line from centre to chord).

    STEP: Plan your approach
    [−1 point ⇒ 2 / 3 points left]

    There are two theorems that you usually use to prove that lines are equal in length:

    • If the angles in a triangle are equal in size, then the sides opposite those angles are equal in length (sides opp equal s).
    • If a line drawn from the centre of a circle is perpendicular to a chord, then it bisects the chord (line from centre to chord).

    We will prove that E^2=90°. Then we will use line from centre to chord to conclude that CE=EB.


    STEP: Determine F^ and hence prove that E^2=90°
    [−1 point ⇒ 1 / 3 points left]

    In Question 1, we proved that B^1=90°x.

    F^=90°x(s in the same seg)

    Now we can use sum of s in ΔECF to determine E^2:

    E^2+x+(90°x)=180° (sum of s in Δ)

    E^2=90°

    STEP: Since FD is a diameter and FDCB, we can conclude that CE=EB
    [−1 point ⇒ 0 / 3 points left]

    CE=EB (line from centre  to chord)

    Choose the right reason! There are three similar theorems about a line from the centre to a chord:

    • To prove that a line from the centre bisects a chord, we must prove that the lines are perpendicular and use line from centre to chord.
    • To prove that a line from the centre is perpendicular to a chord, we must prove that the line is bisected and use line from centre to midpt of chord.
    • To prove that a line passes through the centre of a circle, we must prove that it is a perpendicular bisector of a chord and use perp bisector of chord.

    Submit your answer as: andand

ID is: 4366 Seed is: 7437

Cyclic quads and parallel lines: algebraic questions

Adapted from DBE Nov 2016, Grade 11, P2, Q10
Maths formulas

In the diagram below, P, Q, R, T, and S are points on the circumference of a circle so that RSQT.

RP^Q=2x+43° and SQ^T=x+29°.

  1. Give two simplified expressions for RS^Q.

    Then, choose the reasons for your answers.

    INSTRUCTION: The expressions and reasons can be given in any order. Do not type a degree sign.
    Answer:

    RS^Q=
    OR RS^Q=

    expression
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use geometry theorems to connect RS^Q to one of the angles you have been given. Your answers should have x in them!


    STEP: Use the parallel lines
    [−2 points ⇒ 2 / 4 points left]
    TIP: Whenever we are given parallel lines in a geometry question, we should look out for:
    • alternate angles between parallel lines are equal
    • corresponding angles between parallel lines are equal
    • co-interior angles between parallel lines are supplementary

    We were given RSQT. So, we can use alternate angles between parallel lines:

    RS^Q=x+29° (alt s; RSQT)


    STEP: Use the cyclic quadrilateral
    [−2 points ⇒ 0 / 4 points left]
    TIP: Whenever we are given four points or more on the circumference of a circle, we should look out for cyclic quadrilaterals:
    • opposite angles of a cyclic quadrilateral are supplementary
    • the exterior angle of a cyclic quadrilateral is equal to the opposite interior angle

    PQSR is a cyclic quadrilateral. It has RS^Q and a given angle as opposite interior angles.

    RS^Q+2x+43°=180° (opp s of cyclic quad)

    RS^Q=180°2x43°RS^Q=2x+137°

    Submit your answer as: andand
  2. Hence, or otherwise, determine the size of RP^Q.

    Answer: RP^Q= °
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Can you make an equation which will solve for x?


    STEP: Write and solve an equation for x
    [−2 points ⇒ 0 / 2 points left]

    In Question 1, we found two different expressions which represent RS^Q. Since RS^Q has a fixed value, these two expressions must be equal.

    2x+137°=x+29°108°=3x36°=x

    Now we can substitute the value of x into the expression we were given for RP^Q:

    RP^Q=2(36°)+43°=115°

    Submit your answer as:

ID is: 4366 Seed is: 3806

Cyclic quads and parallel lines: algebraic questions

Adapted from DBE Nov 2016, Grade 11, P2, Q10
Maths formulas

In the diagram below, P, Q, R, T, and S are points on the circumference of a circle so that RSQT.

RP^Q=2x+12° and SQ^T=x12°.

  1. Give two simplified expressions for RS^Q.

    Then, choose the reasons for your answers.

    INSTRUCTION: The expressions and reasons can be given in any order. Do not type a degree sign.
    Answer:

    RS^Q=
    OR RS^Q=

    expression
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use geometry theorems to connect RS^Q to one of the angles you have been given. Your answers should have x in them!


    STEP: Use the parallel lines
    [−2 points ⇒ 2 / 4 points left]
    TIP: Whenever we are given parallel lines in a geometry question, we should look out for:
    • alternate angles between parallel lines are equal
    • corresponding angles between parallel lines are equal
    • co-interior angles between parallel lines are supplementary

    We were given RSQT. So, we can use alternate angles between parallel lines:

    RS^Q=x12° (alt s; RSQT)


    STEP: Use the cyclic quadrilateral
    [−2 points ⇒ 0 / 4 points left]
    TIP: Whenever we are given four points or more on the circumference of a circle, we should look out for cyclic quadrilaterals:
    • opposite angles of a cyclic quadrilateral are supplementary
    • the exterior angle of a cyclic quadrilateral is equal to the opposite interior angle

    PQSR is a cyclic quadrilateral. It has RS^Q and a given angle as opposite interior angles.

    RS^Q+2x+12°=180° (opp s of cyclic quad)

    RS^Q=180°2x12°RS^Q=2x+168°

    Submit your answer as: andand
  2. Hence, or otherwise, determine the size of RP^Q.

    Answer: RP^Q= °
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Can you make an equation which will solve for x?


    STEP: Write and solve an equation for x
    [−2 points ⇒ 0 / 2 points left]

    In Question 1, we found two different expressions which represent RS^Q. Since RS^Q has a fixed value, these two expressions must be equal.

    2x+168°=x12°180°=3x60°=x

    Now we can substitute the value of x into the expression we were given for RP^Q:

    RP^Q=2(60°)+12°=132°

    Submit your answer as:

ID is: 4366 Seed is: 2334

Cyclic quads and parallel lines: algebraic questions

Adapted from DBE Nov 2016, Grade 11, P2, Q10
Maths formulas

In the diagram below, R, Q, S and T are points on the circumference of a circle so that RSQT. QTP is a straight line.

RQ^T=2x+43° and ST^P=x+68°.

  1. Give two simplified expressions for QR^S.

    Then, choose the reasons for your answers.

    INSTRUCTION: The expressions and reasons can be given in any order. Do not type a degree sign.
    Answer:

    QR^S=
    OR QR^S=

    expression
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use geometry theorems to connect QR^S to one of the angles you have been given. Your answers should have x in them!


    STEP: Use the parallel lines
    [−2 points ⇒ 2 / 4 points left]
    TIP: Whenever we are given parallel lines in a geometry question, we should look out for:
    • alternate angles between parallel lines are equal
    • corresponding angles between parallel lines are equal
    • co-interior angles between parallel lines are supplementary

    We were given RSQT. So, we can use co-interior angles between parallel lines:

    QR^S+2x+43°=180° (co-int s; RSQT)

    QR^S=180°2x43°QR^S=2x+137°

    STEP: Use the cyclic quadrilateral
    [−2 points ⇒ 0 / 4 points left]
    TIP: Whenever we are given four points or more on the circumference of a circle, we should look out for cyclic quadrilaterals:
    • opposite angles of a cyclic quadrilateral are supplementary
    • the exterior angle of a cyclic quadrilateral is equal to the opposite interior angle

    QRST is a cyclic quadrilateral. It has an external angle given at ST^P, which is opposite QR^S.

    QR^S=x+68° (ext  of cyclic quad)


    Submit your answer as: andand
  2. Hence, or otherwise, determine the size of RQ^T.

    Answer: RQ^T= °
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Can you make an equation which will solve for x?


    STEP: Write and solve an equation for x
    [−2 points ⇒ 0 / 2 points left]

    In Question 1, we found two different expressions which represent QR^S. Since QR^S has a fixed value, these two expressions must be equal.

    2x+137°=x+68°69°=3x23°=x

    Now we can substitute the value of x into the expression we were given for RQ^T:

    RQ^T=2(23°)+43°=89°

    Submit your answer as:

4. Tangent theorems


ID is: 1619 Seed is: 8647

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line IM¯ which touches the circle at point G. Three points lie on the circle, G,J and K, and five angles are labelled: α,β,θ, etc.

Idenitify an angle which is congruent to angle θ. (If there is no angle congruent to θ, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (IM¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, G: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

The figure below shows the tangent line in red, as well as the angle we are dealing with in the question, angle θ.

In this case, angle θ is not between the tangent line and a chord (it is between the two chords - see above). The tangent-chord theorem does not tell us anything about this angle: it is not connected to a specific angle in the figure.

While there are other angles in the figure which are congruent to each other, we do not know anything useful about angle θ: the correct answer is 'None of the above.'


Submit your answer as:

ID is: 1619 Seed is: 926

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line IL¯ which touches the circle at point K. Three points lie on the circle, K,H and G, and five angles are labelled: 1,2,3, etc.

Idenitify an angle which is congruent to angle 4. (If there is no angle congruent to 4, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (IL¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, K: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

The figure below shows the tangent line in red, as well as the angle we are dealing with in the question, angle 4.

In this case, angle 4 is not between the tangent line and a chord (it is between the two chords - see above). The tangent-chord theorem does not tell us anything about this angle: it is not connected to a specific angle in the figure.

While there are other angles in the figure which are congruent to each other, we do not know anything useful about angle 4: the correct answer is 'None of the above.'


Submit your answer as:

ID is: 1619 Seed is: 6978

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line RW¯ which touches the circle at point P. Three points lie on the circle, P,S and T, and five angles are labelled: 1,2,3, etc.

Idenitify an angle which is congruent to angle 3. (If there is no angle congruent to 3, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (RW¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, P: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle 3

In the figure below, the relevant information is shown in red. It is very helpful to see the figure below as a tangent line and a triangle - the tangent-chord theorem will always include that triangle!

Notice that one of the angles is inside the triangle, while the other angle is outside of the triangle.

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle 4 to the red chord... then from the red chord follow the red arrow across the circle to angle 3 which subtends the chord (angle 3 is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles we just connected are congruent (the same size): angle 3 angle 4. By looking at these two angles, you can see that they do look like they are about the same size; the tangent-chord theorem says that they must be the same size!


Submit your answer as:

ID is: 1615 Seed is: 6165

Applying theorems from Euclidean geometry

The figure below shows a circle with centre O and four points on the circumference of the circle: G,H,I, and J. There are also two angles given: G^=5x4+80° and H^=3x2+40°.

  1. Determine J^ in terms of x. If the answer cannot be determined, write 'No solution'.

    Answer: measure of J^=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    When you read "in terms of x," it means that the answer will be a maths expression which includes the variable x.
    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    There is a lot of information in the question about the diagram, but the diagram itself does not show that information; start by labelling the things that you know. Most specifically, label the information given about any angles in the picture.

    In this question the angles G^ and H^ are given.


    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    The key to this question is the fact that the quadrilateral is cyclic: that means that opposite angles are supplementary. We are trying to find J^, and it must be related to H^ like this: J^+H^=180.

    Substitute in the expression given for H^ and solve for the angle you want: J^+(3x2+40)=180. This works out to be: J^=180(3x2+40)=3x2+140.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As explained in the solution above, the answer to the question comes from the theorem about opposite angles in cyclic quadrilaterals. The correct choice is, 'Opposite angles of a cyclic quadrilateral.'


    Submit your answer as:

ID is: 1615 Seed is: 7418

Applying theorems from Euclidean geometry

The figure below shows a cyclic quadrilateral with vertices at the points A,B,C, and D. There are also two angles given: D^=a3+60° and A^=a230°.

  1. Find B^ in terms of a. If the answer cannot be determined, write 'No solution'.

    Answer: measure of B^=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    When you read "in terms of a," it means that the answer will be a maths expression which includes the variable a.
    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    There is a lot of information in the question about the diagram, but the diagram itself does not show that information; start by labelling the things that you know. Most specifically, label the information given about any angles in the picture. Also, the question states that the quadrilateral is cyclic, so why not sketch a circle around it as a reminder?

    In this question the angles D^ and A^ are given.


    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    The key to this question is the fact that the quadrilateral is cyclic: that means that opposite angles are supplementary. We are trying to find B^, and it must be related to D^ like this: B^+D^=180.

    Substitute in the expression given for D^ and solve for the angle you want: B^+(a3+60)=180. This works out to be: B^=180(a3+60)=a3+120.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As explained in the solution above, the answer to the question comes from the theorem about opposite angles in cyclic quadrilaterals. The correct choice is, 'Opposite angles of a cyclic quadrilateral.'


    Submit your answer as:

ID is: 1615 Seed is: 2813

Applying theorems from Euclidean geometry

The figure below shows a circle with centre O and four points on the circumference of the circle: V,W,X, and Y. There are also two angles given: V^=5x2+75° and W^=x+50°.

  1. Give Y^ in terms of x. If the answer cannot be determined, write 'No solution'.

    Answer: measure of Y^=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    When you read "in terms of x," it means that the answer will be a maths expression which includes the variable x.
    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    There is a lot of information in the question about the diagram, but the diagram itself does not show that information; start by labelling the things that you know. Most specifically, label the information given about any angles in the picture.

    In this question the angles V^ and W^ are given.


    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    The key to this question is the fact that the quadrilateral is cyclic: that means that opposite angles are supplementary. We are trying to find Y^, and it must be related to W^ like this: Y^+W^=180.

    Substitute in the expression given for W^ and solve for the angle you want: Y^+(x+50)=180. This works out to be: Y^=180(x+50)=x+130.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As explained in the solution above, the answer to the question comes from the theorem about opposite angles in cyclic quadrilaterals. The correct choice is, 'Opposite angles of a cyclic quadrilateral.'


    Submit your answer as:

ID is: 4370 Seed is: 1200

Proving the tan chord theorem

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

Use the diagram below to prove the theorem which states that WY^Z=WV^Y, if YZ is a tangent to the circle at Y.

NOTE: There are at least two ways in which we can prove this theorem. However, in this question we have chosen one proof to work through.
  1. Choose a correct construction to prove this theorem. Then choose the correctly worded description to match your construction.

    Answer:

    Correct construction:

    Description:

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    There are at least two possible constructions which can be used to prove the tan chord theorem!

    Think about whether each diagram can be used to connect the two angles that you need to prove equal.

    Could someone follow your instructions to draw the construction exactly, without any more information?


    STEP: Identify the correct construction
    [−0 points ⇒ 1 / 1 points left]

    The following construction can be used to prove the tan chord theorem:

    Construction 2 is wrong because the angle at the centre does not correspond to WV^Y at the circumference.

    Construction 3 is wrong because the line YB is not a diameter and so YB does not meet the tangent YZ at 90°.


    STEP: Describe the construction fully
    [−1 point ⇒ 0 / 1 points left]

    The written description of a construction should provide all the details needed to draw the construction with no extra information.

    Construct OV and OY.
    This is an incorrect construction.
    Construct radii.
    It does not say which radii to draw.
    Construct BY and BV.
    It does not say that BY has to be a diameter.
    Construct radii OY and OW.

    Submit your answer as:
    and
  2. Complete the proof by selecting the options below.

    Answer:

    Let Y1=x

    Statement Reason
    Y2+Y1=90°
    Y2=
    OY=OW
    W2=x
    YOW=180°2x (sum of s in Δ)
    V= ( at centre =2× at circumference)

    V=Y2

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use circle geometry theorems to write other relevant angles in terms of x.


    STEP: Complete the proof
    [−4 points ⇒ 0 / 4 points left]
    TIP: You must learn, and use, the correct geometry reasons wording for proofs.

    Construct radii OY and OW.

    Let Y1=x

    Statement Reason
    Y2+Y1=90° (tan radius)
    Y2=90°x
    OY=OW (radii)
    W2=x (s opp equal sides)
    YOW=180°2x (sum of s in Δ)
    V= 90°x ( at centre =2× at circumference)

    V=Y2


    Submit your answer as: andandandand

ID is: 4370 Seed is: 8965

Proving the tan chord theorem

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

Use the diagram below to prove the theorem which states that QR^S=QP^R, if RS is a tangent to the circle at R.

NOTE: There are at least two ways in which we can prove this theorem. However, in this question we have chosen one proof to work through.
  1. Choose a correct construction to prove this theorem. Then choose the correctly worded description to match your construction.

    Answer:

    Correct construction:

    Description:

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    There are at least two possible constructions which can be used to prove the tan chord theorem!

    Think about whether each diagram can be used to connect the two angles that you need to prove equal.

    Could someone follow your instructions to draw the construction exactly, without any more information?


    STEP: Identify the correct construction
    [−0 points ⇒ 1 / 1 points left]

    The following construction can be used to prove the tan chord theorem:

    Construction 2 is wrong because the angle at the centre does not correspond to QP^R at the circumference.

    Construction 3 is wrong because the line RV is not a diameter and so RV does not meet the tangent RS at 90°.


    STEP: Describe the construction fully
    [−1 point ⇒ 0 / 1 points left]

    The written description of a construction should provide all the details needed to draw the construction with no extra information.

    Construct VR and VP.
    It does not say that VR has to be a diameter.
    Construct diameter VOR. Join V to P.
    Construct diameter.
    It does not say which diameter to construct, or that it must be joined up with P.
    Construct OP and OR.
    This is an incorrect construction.

    Submit your answer as:
    and
  2. Complete the proof by selecting the options below.

    Answer:

    Let R1=x

    Statement Reason
    R1+R2=90°
    R2=
    P1=x
    P1+P2=90°
    P2=

    P2=R2

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use circle geometry theorems to write other relevant angles in terms of x.


    STEP: Complete the proof
    [−4 points ⇒ 0 / 4 points left]
    TIP: You must learn, and use, the correct geometry reasons wording for proofs.

    Construct diameter VOR. Join V to P.

    Let R1=x

    Statement Reason
    R1+R2=90° (tan diameter)
    R2=90°x
    P1=x (s in the same seg)
    P1+P2=90° ( in semi circle)
    P2=90°x

    P2=R2


    Submit your answer as: andandandand

ID is: 4370 Seed is: 5566

Proving the tan chord theorem

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

Use the diagram below to prove the theorem which states that EF^G=ED^F, if FG is a tangent to the circle at F.

NOTE: There are at least two ways in which we can prove this theorem. However, in this question we have chosen one proof to work through.
  1. Choose a correct construction to prove this theorem. Then choose the correctly worded description to match your construction.

    Answer:

    Correct construction:

    Description:

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    There are at least two possible constructions which can be used to prove the tan chord theorem!

    Think about whether each diagram can be used to connect the two angles that you need to prove equal.

    Could someone follow your instructions to draw the construction exactly, without any more information?


    STEP: Identify the correct construction
    [−0 points ⇒ 1 / 1 points left]

    The following construction can be used to prove the tan chord theorem:

    Construction 1 is wrong because the angle at the centre does not correspond to ED^F at the circumference.

    Construction 3 is wrong because the line FC is not a diameter and so FC does not meet the tangent FG at 90°.


    STEP: Describe the construction fully
    [−1 point ⇒ 0 / 1 points left]

    The written description of a construction should provide all the details needed to draw the construction with no extra information.

    Construct OD and OF.
    This is an incorrect construction.
    Construct CF and CD.
    It does not say that CF has to be a diameter.
    Construct radii.
    It does not say which radii to draw.
    Construct radii OF and OE.

    Submit your answer as:
    and
  2. Complete the proof by selecting the options below.

    Answer:

    Let F1=x

    Statement Reason
    (tan radius)
    F2= 90°x
    OF=OE
    E2=x
    FOE=180°2x
    D= ( at centre =2× at circumference)

    D=F2

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use circle geometry theorems to write other relevant angles in terms of x.


    STEP: Complete the proof
    [−4 points ⇒ 0 / 4 points left]
    TIP: You must learn, and use, the correct geometry reasons wording for proofs.

    Construct radii OF and OE.

    Let F1=x

    Statement Reason
    F2+F1=90° (tan radius)
    F2=90°x
    OF=OE (radii)
    E2=x (s opp equal sides)
    FOE=180°2x (sum of s in Δ)
    D= 90°x ( at centre =2× at circumference)

    D=F2


    Submit your answer as: andandandand

ID is: 4357 Seed is: 5681

Tangents and radii

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle KL. J is a point on chord LK such that OJKL. LM and KM are tangents to the circle at L and K respectively.

LM=15units; JO^L=52.5°

Calculate the size of MK^L. Choose two of the reasons you used to justify your answer from the options provided.

INSTRUCTION: Round your answer to one decimal place.
Answer: MK^L= °
Two of my reasons:

one-of
type(numeric.abserror(0.1))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You have been given two tangents, so you should be able to use one or more of the tangent theorems.

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

STEP: Find OL^J
[−1 point ⇒ 4 / 5 points left]

In ΔJOL,

OL^J=180°90°52.5° (sum of s in Δ)

OL^J=37.5°


STEP: Find KL^M
[−2 points ⇒ 2 / 5 points left]

We have been given two tangents, LM and KM. So, we recall the theorems about tangents:

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

Since LM is a tangent, and OL is a radius,

OL^M=90° (tan  radius)

OL^M=KL^M+OL^J90°=KL^M+37.5°KL^M=52.5°

STEP: Find MK^L
[−2 points ⇒ 0 / 5 points left]

Since M is a common point for tangents LM and KM,

LM=KM (tans from same pt)

in ΔKLM:

MK^L=52.5° (s opp equal sides)


Submit your answer as: andand

ID is: 4357 Seed is: 1693

Tangents and radii

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle QR. P is a point on chord RQ such that OPQR. RS and QS are tangents to the circle at R and Q respectively.

RS=15units; SQ^R=59.7°

Calculate the size of PO^R. Choose two of the reasons you used to justify your answer from the options provided.

INSTRUCTION: Round your answer to one decimal place.
Answer: PO^R= °
Two of my reasons:

one-of
type(numeric.abserror(0.1))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You have been given two tangents, so you should be able to use one or more of the tangent theorems.

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

STEP: Find QR^S
[−2 points ⇒ 3 / 5 points left]

We have been given two tangents, RS and QS. So, we recall the theorems about tangents:

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

Since S is a common point for tangents RS and QS,

RS=QS (tans from same pt)

in ΔQRS:

QR^S=59.7° (s opp equal sides)


STEP: Find OR^P
[−2 points ⇒ 1 / 5 points left]

Since RS is a tangent, and OR is a radius,

OR^S=90° (tan  radius)

OR^S=QR^S+OR^P90°=59.7°+OR^POR^P=30.3°

STEP: Find PO^R
[−1 point ⇒ 0 / 5 points left]

In ΔPOR,

PO^R=180°90°30.3° (sum of s in Δ)

PO^R=59.7°


Submit your answer as: andand

ID is: 4357 Seed is: 2623

Tangents and radii

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle QR. P is a point on chord RQ such that OPQR. RS and QS are tangents to the circle at R and Q respectively.

RS=15units; SQ^R=52.7°

Calculate the size of PO^R. Choose two of the reasons you used to justify your answer from the options provided.

INSTRUCTION: Round your answer to one decimal place.
Answer: PO^R= °
Two of my reasons:

one-of
type(numeric.abserror(0.1))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You have been given two tangents, so you should be able to use one or more of the tangent theorems.

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

STEP: Find QR^S
[−2 points ⇒ 3 / 5 points left]

We have been given two tangents, RS and QS. So, we recall the theorems about tangents:

  • The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact.
  • Two tangents drawn to a circle from the same point outside the circle are equal in length.
  • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

Since S is a common point for tangents RS and QS,

RS=QS (tans from same pt)

in ΔQRS:

QR^S=52.7° (s opp equal sides)


STEP: Find OR^P
[−2 points ⇒ 1 / 5 points left]

Since RS is a tangent, and OR is a radius,

OR^S=90° (tan  radius)

OR^S=QR^S+OR^P90°=52.7°+OR^POR^P=37.3°

STEP: Find PO^R
[−1 point ⇒ 0 / 5 points left]

In ΔPOR,

PO^R=180°90°37.3° (sum of s in Δ)

PO^R=52.7°


Submit your answer as: andand

ID is: 1621 Seed is: 578

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment CG¯ touching the circle at point A. There are four points on the circle: A,B,D and E, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles has the same measure as angle ψ? If there are no angles congruent to ψ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle ψ

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle δ to the red chord... then from the red chord follow the red arrow across the circle to angle ψ which subtends the chord (angle ψ is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): ψ δ.


Submit your answer as:

ID is: 1621 Seed is: 4884

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment BE¯ touching the circle at point D. There are four points on the circle: D,C,A and G, and the following angles are labelled: 1,2,3, etc. (The diagram is drawn to scale.)

Which of the angles has the same measure as angle 7? If there are no angles congruent to 7, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle 7

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect these pieces, move through the figure like this: from the tangent line, rotate through angle 6 and angle 5 up to chord DA¯ - you need to ignore DC¯! Then from the red chord go across the triangle to angle 7 which subtends the chord (angle 7 is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): angle 7 angle (5+6).


Submit your answer as:

ID is: 1621 Seed is: 5934

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment IL¯ touching the circle at point K. There are four points on the circle: K,J,H and G, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles has the same measure as angle χ? If there are no angles congruent to χ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord - in this question it is angle χ
  4. another angle subtending the chord

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle χ to the red chord... then from the red chord follow the red arrow across the circle to angle β which subtends the chord (angle β is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): χ β.


Submit your answer as:

ID is: 3934 Seed is: 5634

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, PQRT is a cyclic quadrilateral having PTQR. The tangent at Q meets PT produced at S. QR=PR and QP^S=68°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why Q^2=68°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that PQ is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines PT and QR. The line PQ is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, Q^2 and P^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    Q^2=P^1

    P^1 is given in the question as 68°.

    Q^2=P^1Q2=68°

    Q^2 must be equal to 68° because alternate interior angles are equal. The correct answer is Option B.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. R^1
    2. P^3
    Answer:
    1. R^1= °
      Reason:
    2. P^3= °
      Reason:
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find R^1 using the fact that triangle PQR is isosceles.


    STEP: Find R^1 using the fact that triangle PQR is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that Q^2=68°. We also know from the question that triangle PQR is isosceles with Q^2 equal to P^2. So that means P^2 must also be 68°.

    Now we know two of the angles in triangle PQR so we can calculate the size of R^1:

    R^1=180°2(68°)=44°

    To find both angles in triangle PQR, we used the sum of angles in a triangle.


    STEP: Find P^3 using angles on a straight line
    [−2 points ⇒ 0 / 5 points left]

    At Point P there are three angles, and they make a straight line. So they must have a sum of 180°.

    180°=P^1+P^2+P^3

    We know from the diagram that P^1=68°. And from the work above we know that P^2 is also equal to 68°.

    180°=(68°)+(68°)+P^344°=P^3
    NOTE: We can also get this answer using alternate interior angles (on parallel lines) with R^1. But even though that is an accurate reason, it is not one of the answer options.

    The key reason for this information is angles on a straight line.


    Submit your answer as: andandand

ID is: 3934 Seed is: 9601

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, ABCE is a cyclic quadrilateral having AEBC. The tangent at B meets AE produced at D. BC=AC and BA^D=70°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why B^2=70°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that AB is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines AE and BC. The line AB is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, B^2 and A^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    B^2=A^1

    A^1 is given in the question as 70°.

    B^2=A^1B2=70°

    B^2 must be equal to 70° because alternate interior angles are equal. The correct answer is Option B.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. C^1
    2. A^3
    Answer:
    1. C^1= °
      Reason:
    2. A^3= °
      Reason:
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find C^1 using the fact that triangle ABC is isosceles.


    STEP: Find C^1 using the fact that triangle ABC is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that B^2=70°. We also know from the question that triangle ABC is isosceles with B^2 equal to A^2. So that means A^2 must also be 70°.

    Now we know two of the angles in triangle ABC so we can calculate the size of C^1:

    C^1=180°2(70°)=40°

    To find both angles in triangle ABC, we used the sum of angles in a triangle.


    STEP: Find A^3 using angles on a straight line
    [−2 points ⇒ 0 / 5 points left]

    At Point A there are three angles, and they make a straight line. So they must have a sum of 180°.

    180°=A^1+A^2+A^3

    We know from the diagram that A^1=70°. And from the work above we know that A^2 is also equal to 70°.

    180°=(70°)+(70°)+A^340°=A^3
    NOTE: We can also get this answer using alternate interior angles (on parallel lines) with C^1. But even though that is an accurate reason, it is not one of the answer options.

    The key reason for this information is angles on a straight line.


    Submit your answer as: andandand

ID is: 3934 Seed is: 9727

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, PQRT is a cyclic quadrilateral having PTQR. The tangent at Q meets PT produced at S. QR=PR and QP^S=67°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why Q^2=67°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that PQ is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines PT and QR. The line PQ is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, Q^2 and P^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    Q^2=P^1

    P^1 is given in the question as 67°.

    Q^2=P^1Q2=67°

    Q^2 must be equal to 67° because alternate interior angles are equal. The correct answer is Option A.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. R^1
    2. P^3
    Answer:
    1. R^1= °
      Reason:
    2. P^3= °
      Reason:
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find R^1 using the fact that triangle PQR is isosceles.


    STEP: Find R^1 using the fact that triangle PQR is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that Q^2=67°. We also know from the question that triangle PQR is isosceles with Q^2 equal to P^2. So that means P^2 must also be 67°.

    Now we know two of the angles in triangle PQR so we can calculate the size of R^1:

    R^1=180°2(67°)=46°

    To find both angles in triangle PQR, we used the sum of angles in a triangle.


    STEP: Find P^3 using angles on a straight line
    [−2 points ⇒ 0 / 5 points left]

    At Point P there are three angles, and they make a straight line. So they must have a sum of 180°.

    180°=P^1+P^2+P^3

    We know from the diagram that P^1=67°. And from the work above we know that P^2 is also equal to 67°.

    180°=(67°)+(67°)+P^346°=P^3
    NOTE: We can also get this answer using alternate interior angles (on parallel lines) with R^1. But even though that is an accurate reason, it is not one of the answer options.

    The key reason for this information is angles on a straight line.


    Submit your answer as: andandand

ID is: 1622 Seed is: 5499

Using the tangent-chord theorem to find angles

Line segment BE¯ is tangent to circle O at point D. Points A and G lie on the circle. Two angles are given: ED^G=75° and BD^A=75°.

Work out the values of the three angles in the triangle, α,β and θ.

Answer:

α=°
β=°
θ=°

numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle β, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle β
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: β makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

75°+β+75°=180°150°+β=180°β=30°

Cool: we have one of the three answers: β=30°.


STEP: Connect information using the tangent-chord theorem to find α
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both θ and α to angles that we know. For example, let's look at the 75° angle on the left of the tangent point: it reaches from the tangent line to chord DG¯. From the chord, we find angle α across the triangle. This is the angle which must be congruent to the 75° angle. (See the green parts of the figure below.) So α=75°.


STEP: Repeat the previous step to find θ
[−1 point ⇒ 0 / 3 points left]

The other 75° angle and θ are related in exactly the same way. The diagram shows the connection between these angles in purple. So θ=75°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: α=75°,β=30° and θ=75°.


Submit your answer as: andand

ID is: 1622 Seed is: 6557

Using the tangent-chord theorem to find angles

Line segment IL¯ is tangent to circle O at point K. Points H and G lie on the circle. Two angles are given: LK^G=75° and IK^H=75°.

Determine the values of the three angles in the triangle, x,y and z.

Answer:

x=°
y=°
z=°

numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle y, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle y
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: y makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

75°+y+75°=180°150°+y=180°y=30°

Cool: we have one of the three answers: y=30°.


STEP: Connect information using the tangent-chord theorem to find x
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both z and x to angles that we know. For example, let's look at the 75° angle on the left of the tangent point: it reaches from the tangent line to chord KG¯. From the chord, we find angle x across the triangle. This is the angle which must be congruent to the 75° angle. (See the green parts of the figure below.) So x=75°.


STEP: Repeat the previous step to find z
[−1 point ⇒ 0 / 3 points left]

The other 75° angle and z are related in exactly the same way. The diagram shows the connection between these angles in purple. So z=75°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: x=75°,y=30° and z=75°.


Submit your answer as: andand

ID is: 1622 Seed is: 1444

Using the tangent-chord theorem to find angles

Line segment IM¯ is tangent to circle O at point G. Points J and K lie on the circle. Two angles are given: MG^K=57° and IG^J=85°.

Work out the values of the three angles in the triangle, a,b and c.

Answer:

a=°
b=°
c=°

numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle a, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle a
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: a makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

57°+a+85°=180°142°+a=180°a=38°

Cool: we have one of the three answers: a=38°.


STEP: Connect information using the tangent-chord theorem to find b
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both c and b to angles that we know. For example, let's look at the 57° angle : it reaches from the tangent line to chord GK¯. From the chord, we find angle b across the triangle. This is the angle which must be congruent to the 57° angle. (See the green parts of the figure below.) So b=57°.


STEP: Repeat the previous step to find c
[−1 point ⇒ 0 / 3 points left]

The 85° angle and c are related in exactly the same way. The diagram shows the connection between these angles in purple. So c=85°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: a=38°,b=57° and c=85°.


Submit your answer as: andand

ID is: 1620 Seed is: 2712

Two tangents from a single point

In the figure below, two line segments go from point C to a circle. Both segments are tangent to the circle: the tangent points are A and B. The centre of the circle is at point O and the radius of the circle is 5.1 cm. Line segment AC¯ is 10 cm long. The length of AD¯ is 4.5 cm. (The figure is drawn to scale!)

Find the length of BC¯ and select what type of angle CA^D is (right angle, obtuse, acute, etc.).

Answer: BC¯= cm
CA^D is
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point C must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, BC¯AC¯, which means that BC¯=10 cm.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle CA^D, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point A.

Is CA^D larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

The angle CA^D is smaller than the right angle: the correct answer is 'acute.'


Submit your answer as: and

ID is: 1620 Seed is: 2149

Two tangents from a single point

In the figure below, two line segments go from point I to a circle. Both segments are tangent to the circle: the tangent points are G and H. The centre of the circle is at point O and the radius of the circle is 6.6 cm. Line segment GI¯ is 13 cm long. The length of GJ¯ is 11.2 cm. (The figure is drawn to scale!)

Find the length of HI¯ and select what type of angle IG^J is (right angle, obtuse, acute, etc.).

Answer: HI¯= cm
IG^J is
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point I must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, HI¯GI¯, which means that HI¯=13 cm.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle IG^J, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point G.

Is IG^J larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

You can see in the figure that IG^J is larger than the right angle. Therefore, the angle is obtuse.


Submit your answer as: and

ID is: 1620 Seed is: 1940

Two tangents from a single point

In the figure below, two line segments go from point R to a circle. Both segments are tangent to the circle: the tangent points are P and Q. The centre of the circle is at point O and the radius of the circle is 17.9 cm. Line segment PR¯ is 35 cm long. The length of QS¯ is 16.5 cm. (The figure is drawn to scale!)

Determine the length of QR¯ and select what type of angle RQ^S is (right angle, obtuse, acute, etc.).

Answer: QR¯= cm
RQ^S is
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point R must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, QR¯PR¯, which means that QR¯=35 cm.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle RQ^S, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point Q.

Is RQ^S larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

The angle RQ^S is smaller than the right angle: the correct answer is 'acute.'


Submit your answer as: and

ID is: 1614 Seed is: 1930

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Points G and H are on the circumference of the circle. There is a single point labelled outside of the circle. There is a tangent line at point G on the circle.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    In this case, there is one point outside the circle: point I. Does it have two lines coming from it or only one? Are the lines tangent to the circle or not?

    While there is a tangent line in this figure (tangent to the circle at point G), there are not two congruent line segments.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: none.


    Submit your answer as:
  2. Identify one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? There is a radius shown, but it does not meet a tangent line!

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

ID is: 1614 Seed is: 5905

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. There are four points on the circle: K,J,H and G. There is a single point labelled outside of the circle. There are two lines tangent to the circle: they touch the circle at points K and J.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    In this case, there is one point outside the circle: point I. Does it have two lines coming from it or only one? Are the lines tangent to the circle or not?

    In this case, the diagram does include a pair of tangent lines: they pass from point I and are tangent to the circle at points K and J. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: IJ;IK.


    Submit your answer as:
  2. Name one right angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no right angles in the figure, write none.
    Answer:
    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? The figure below shows point O as a red dot... and there is no radius in the picture at all. Therefore we cannot know if there is a right angle in the diagram.

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

ID is: 1614 Seed is: 2449

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Three points lie on the circle: A,B and D. There is a single point labelled outside of the circle. There are two lines tangent to the circle: they touch the circle at points A and B.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    In this case, there is one point outside the circle: point C. Does it have two lines coming from it or only one? Are the lines tangent to the circle or not?

    In this case, the diagram does include a pair of tangent lines: they pass from point C and are tangent to the circle at points A and B. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: BC;AC.


    Submit your answer as:
  2. Name one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? The figure below shows point O as a red dot... and there is no radius in the picture at all. Therefore we cannot know if there is a right angle in the diagram.

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

5. Mixed applications


ID is: 4412 Seed is: 3345

Proving parallel lines using right angles

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, KOL is a diameter of the circle with centre O. JP is a tangent to the circle at J and JN=NL.

Prove that OMKJ.

Answer: KJ^L= °
N^2= °
OMKJ
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

STEP: Plan your approach
[−0 points ⇒ 4 / 4 points left]
TIP:

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

We will use the information given in the diagram to find a pair of angles which satisfies one of these relationships.


STEP: Use the fact that KOL is a diameter
[−1 point ⇒ 3 / 4 points left]

KOL is a diameter. It subtends KJ^L at the circumference.

KJ^L=90° ( in semi circle)


STEP: Use the fact that JN=NL
[−2 points ⇒ 1 / 4 points left]

ON is a line from the centre. It goes to the midpoint of chord JL.

N^2=90° (line from centre to midpt of chord)


STEP: Prove that OMKJ
[−1 point ⇒ 0 / 4 points left]
KJ^L=N^2=90°

OMKJ (corresp s equal)

TIP: Any of the three theorems (alternate, corresponding, or co-interior) could have been used to prove that OMKJ in this question. But, we have shown that N^2=KJ^L so we must choose "corresp s equal" as the correct reason.

Submit your answer as: andandandand

ID is: 4412 Seed is: 2343

Proving parallel lines using right angles

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, WOX is a diameter of the circle with centre O. VP is a tangent to the circle at V and VZ=ZX.

Prove that OYWV.

Answer: WV^X= °
Z^2= °
OYWV
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

STEP: Plan your approach
[−0 points ⇒ 4 / 4 points left]
TIP:

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

We will use the information given in the diagram to find a pair of angles which satisfies one of these relationships.


STEP: Use the fact that WOX is a diameter
[−1 point ⇒ 3 / 4 points left]

WOX is a diameter. It subtends WV^X at the circumference.

WV^X=90° ( in semi circle)


STEP: Use the fact that VZ=ZX
[−2 points ⇒ 1 / 4 points left]

OZ is a line from the centre. It goes to the midpoint of chord VX.

Z^2=90° (line from centre to midpt of chord)


STEP: Prove that OYWV
[−1 point ⇒ 0 / 4 points left]
WV^X=Z^2=90°

OYWV (corresp s equal)

TIP: Any of the three theorems (alternate, corresponding, or co-interior) could have been used to prove that OYWV in this question. But, we have shown that Z^2=WV^X so we must choose "corresp s equal" as the correct reason.

Submit your answer as: andandandand

ID is: 4412 Seed is: 5510

Proving parallel lines using right angles

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, BOC is a diameter of the circle with centre O. AP is a tangent to the circle at A and AE=EC.

Prove that ODBA.

Answer: BA^C= °
E^2= °
ODBA
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

STEP: Plan your approach
[−0 points ⇒ 4 / 4 points left]
TIP:

There are three ways to prove that lines are parallel:

  • alternate angles are equal
  • corresponding angles are equal
  • co-interior angles are supplementary

We will use the information given in the diagram to find a pair of angles which satisfies one of these relationships.


STEP: Use the fact that BOC is a diameter
[−1 point ⇒ 3 / 4 points left]

BOC is a diameter. It subtends BA^C at the circumference.

BA^C=90° ( in semi circle)


STEP: Use the fact that AE=EC
[−2 points ⇒ 1 / 4 points left]

OE is a line from the centre. It goes to the midpoint of chord AC.

E^2=90° (line from centre to midpt of chord)


STEP: Prove that ODBA
[−1 point ⇒ 0 / 4 points left]
BA^C=E^2=90°

ODBA (corresp s equal)

TIP: Any of the three theorems (alternate, corresponding, or co-interior) could have been used to prove that ODBA in this question. But, we have shown that E^2=BA^C so we must choose "corresp s equal" as the correct reason.

Submit your answer as: andandandand

ID is: 4411 Seed is: 242

Proving tangents using cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, O is the centre of the circle. VP is a tangent to the circle at V and WZ=ZX.

Ndidi has already proven that VPWX and that VOXP is a cyclic quad.

Help Ndidi to prove that XP is a tangent to the circle at X by choosing from the options provided.

INSTRUCTION: Make sure that your proof makes sense. True statements that do not help to prove that XP is a tangent will still be marked wrong.
Answer: =
=
XP is a tangent to the circle at Point X
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways that we can prove that a line is tangent to a circle:

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

STEP: Plan how to prove that XP is a tangent
[−1 point ⇒ 3 / 4 points left]
TIP:

There are two ways that we can prove that a line is a tangent to a circle.

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

So, we need to prove that OX^P=90°.

We must use other theorems, which don’t involve XP being a tangent, to prove the fact above.


STEP: Prove that XP is a tangent
[−3 points ⇒ 0 / 4 points left]

VP is a tangent to the circle and OV is a radius. So,

OV^P=90° (tan  radius)

OX^P=90° (opp s of cyclic quad)

XP is a tangent to the circle at Point X (converse tan  radius)

NOTE: There is often more than one way of proving something in Geometry. Any correct proof would be marked right by your teacher. However, in this question we asked you to prove it in a specific way.

Submit your answer as: andandandandandand

ID is: 4411 Seed is: 3858

Proving tangents using cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, O is the centre of the circle. JP is a tangent to the circle at J and KN=NL.

Mandilakhe has already proven that JPKL and that JOLP is a cyclic quad.

Help Mandilakhe to prove that LP is a tangent to the circle at L by choosing from the options provided.

INSTRUCTION: Make sure that your proof makes sense. True statements that do not help to prove that LP is a tangent will still be marked wrong.
Answer: =
=
LP is a tangent to the circle at Point L
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways that we can prove that a line is tangent to a circle:

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

STEP: Plan how to prove that LP is a tangent
[−1 point ⇒ 3 / 4 points left]
TIP:

There are two ways that we can prove that a line is a tangent to a circle.

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

So, we need to prove that OL^P=90°.

We must use other theorems, which don’t involve LP being a tangent, to prove the fact above.


STEP: Prove that LP is a tangent
[−3 points ⇒ 0 / 4 points left]

JP is a tangent to the circle and OJ is a radius. So,

OJ^P=90° (tan  radius)

OL^P=90° (opp s of cyclic quad)

LP is a tangent to the circle at Point L (converse tan  radius)

NOTE: There is often more than one way of proving something in Geometry. Any correct proof would be marked right by your teacher. However, in this question we asked you to prove it in a specific way.

Submit your answer as: andandandandandand

ID is: 4411 Seed is: 3456

Proving tangents using cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, O is the centre of the circle. JP is a tangent to the circle at J and KN=NL.

Ntuthuko has already proven that JPKL and that JOLP is a cyclic quad.

Help Ntuthuko to prove that LP is a tangent to the circle at L by choosing from the options provided.

INSTRUCTION: Make sure that your proof makes sense. True statements that do not help to prove that LP is a tangent will still be marked wrong.
Answer: =
=
LP is a tangent to the circle at Point L
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways that we can prove that a line is tangent to a circle:

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

STEP: Plan how to prove that LP is a tangent
[−1 point ⇒ 3 / 4 points left]
TIP:

There are two ways that we can prove that a line is a tangent to a circle.

  • If a line is drawn through the end-point of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle: (converse tan chord theorem).
  • If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle: (converse tan radius).

So, we need to prove that OL^P=90°.

We must use other theorems, which don’t involve LP being a tangent, to prove the fact above.


STEP: Prove that LP is a tangent
[−3 points ⇒ 0 / 4 points left]

JP is a tangent to the circle and OJ is a radius. So,

OJ^P=90° (tan  radius)

OL^P=90° (opp s of cyclic quad)

LP is a tangent to the circle at Point L (converse tan  radius)

NOTE: There is often more than one way of proving something in Geometry. Any correct proof would be marked right by your teacher. However, in this question we asked you to prove it in a specific way.

Submit your answer as: andandandandandand

ID is: 4376 Seed is: 6835

Euclidean proofs

Adapted from DBE Nov 2015 Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices A, C and F of ΔACF are concyclic. BC and BF are tangents to the circle at C and F respectively. T is a point on AC such that TBAF. CF cuts TB at E.

  1. Prove that C^1=T^3.

    INSTRUCTION: Complete the proof as it is set out here. Read through the whole proof before you start.
    Answer:
    Statement Reason
    C1=
    Also,
    C1=T3

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Start with C^1 and see if you can find other angles that are equal to this angle. Use only the given information.


    STEP: Use tan chord theorem
    [−2 points ⇒ 2 / 4 points left]

    We are going to start with C^1 and find other angles that are equal to this angle. We will use only the given information.

    C^1=A^ (tan chord theorem)


    STEP: Use the fact that TBAF
    [−2 points ⇒ 0 / 4 points left]

    Also, A^=T^3 (corresp s; TBAF)

    C^1=T^3
    TIP: Your statements and reasons must work together to prove that C^1=T^3. It is not enough to just choose options that are true.

    Submit your answer as: andandand
  2. Prove that TCBF is a cyclic quadrilateral.

    Here is the diagram again:

    Answer:
    Statement Reason
    CB=FB
    (s opp equal sides)
    TCBF is a cyclic quad
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    There are three theorems that we use to prove that a quadrilateral is cyclic:

    1. If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).
    2. If the exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is cyclic (converse ext of cyclic quad).
    3. If a line segment subtends equal angles at two points on the same side of the line segment, then the four points are concyclic (i.e. they form a cyclic quad) (converse s in the same seg).

    STEP: Plan your approach
    [−1 point ⇒ 3 / 4 points left]

    There are three theorems that we use to prove that a quadrilateral is cyclic:

    1. If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).
    2. If the exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is cyclic (converse ext of cyclic quad).
    3. If a line segment subtends equal angles at two points on the same side of the line segment, then the four points are concyclic (i.e. they form a cyclic quad) (converse s in the same seg).

    We have the right information to use the last theorem.


    STEP: Use tans from same pt and s opp equal sides
    [−2 points ⇒ 1 / 4 points left]

    CB=FB (tans from same pt)

    C^1=F^3 (s opp equal sides)

    But, C^1=T^3 (proven above)

    F^3=T^3


    STEP: Use converse s in the same seg
    [−1 point ⇒ 0 / 4 points left]

    TCBF is a cyclic quad (converse s in the same seg)


    Submit your answer as: andandand
  3. Prove that BT bisects CT^F.

    Here is the diagram again:

    Answer:
    Statement Reason
    T2=
    But, T3=
    T2=T3
    Therefore, BT bisects CTF.
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    "Bisect" means to cut into two equal pieces.


    STEP: Use what you have proven already
    [−2 points ⇒ 0 / 2 points left]
    "Bisect" means to cut into two equal pieces.

    We need to prove that T^2=T^3.

    TIP: We have proven that TCBF is a cyclic quad. So we can use s in the same seg.

    T^2=C^1 (s in the same seg)

    But, C^1=T^3 (proven above)

    T^2=T^3

    Therefore, BT bisects CT^F.


    Submit your answer as: andandand

ID is: 4376 Seed is: 6129

Euclidean proofs

Adapted from DBE Nov 2015 Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices B, F and E of ΔBFE lie on the circumference of the circle. CF and CE are tangents to the circle at F and E respectively.

It is given that TECF is a cyclic quadrilateral.

  1. Prove that E^3=T^2.

    INSTRUCTION: Complete the proof as it is set out here. Read through the whole proof before you start.
    Answer:
    Statement Reason
    E3=F1 (s opp equal sides)
    Also,
    E3=T2

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Start with E^3 and see if you can find other angles that are equal to this angle. Use only the given information.


    STEP: Use s opp equal sides
    [−2 points ⇒ 2 / 4 points left]

    We are going to start with E^3 and find other angles that are equal to this angle. We will use only the given information.

    FC=EC (tans from same pt)

    E^3=F^1 (s opp equal sides)


    STEP: Use s in the same seg
    [−2 points ⇒ 0 / 4 points left]

    Also, F^1=T^2 (s in the same seg)

    E^3=T^2
    TIP: We were given that TFCE is a cyclic quad. That is why we can use s in the same seg.

    Submit your answer as: andandand
  2. Prove that TCBE.

    Here is the diagram again:

    Answer:
    Statement Reason
    B=E3
    Also, E3=T3
    TCBE
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    There are three theorems that we usually use to prove that lines are parallel:

    1. If the corresponding angles on two lines are equal, then the lines are parallel (corresp s equal).
    2. If the alternate angles on two lines are equal, then the lines are parallel (alt s equal).
    3. If the co-interior angles on two lines are supplementary, then the lines are parallel (co-int s supp).

    STEP: Plan your approach
    [−1 point ⇒ 3 / 4 points left]

    There are three theorems that we usually use to prove that lines are parallel:

    1. If the corresponding angles on two lines are equal, then the lines are parallel (corresp s equal).
    2. If the alternate angles on two lines are equal, then the lines are parallel (alt s equal).
    3. If the co-interior angles on two lines are supplementary, then the lines are parallel (co-int s supp).

    We have the right information to use the first theorem.


    STEP: Use tan chord theorem and s in the same seg
    [−2 points ⇒ 1 / 4 points left]

    B^=E^3 (tan chord theorem)

    Also, E^3=T^3 (s in the same seg)

    B^=T^3


    STEP: Use corresp s equal
    [−1 point ⇒ 0 / 4 points left]

    TCBE (corresp s equal)

    Choose the right reason to prove that the lines are parallel:

    corresp s equal is used to prove that lines are parallel.

    corresp s; TCBE is used when we know that lines are parallel and we want to prove that angles are equal.


    Submit your answer as: andandand
  3. Given that T is the centre of the circle, prove that TC bisects FC^E.

    Here is the diagram again:

    Answer:
    Statement Reason
    C1=F2
    Similarly,
    But TF=TE
    F2=E2
    C1=C2
    TC bisects FCE
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    "Bisect" means to cut into two equal pieces.


    STEP: Prove that C^1=C^2 using Euclidean geometry theorems
    [−2 points ⇒ 0 / 2 points left]
    "Bisect" means to cut into two equal pieces.

    C^1=F^2 (s in the same seg)

    Similarly, C^2=E^2 (s in the same seg)

    But TF=TE (radii)

    F^2=E^2 (s opp equal sides)

    C^1=C^2

    Therefore, TC bisects FC^E.


    Submit your answer as: andandand

ID is: 4376 Seed is: 8934

Euclidean proofs

Adapted from DBE Nov 2015 Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices B, F and C of ΔBFC are concyclic. AF and AC are tangents to the circle at F and C respectively. T is a point on BF such that TABC. FC cuts TA at E.

  1. Prove that F^1=T^3.

    INSTRUCTION: Complete the proof as it is set out here. Read through the whole proof before you start.
    Answer:
    Statement Reason
    F1=
    Also,
    F1=T3

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Start with F^1 and see if you can find other angles that are equal to this angle. Use only the given information.


    STEP: Use tan chord theorem
    [−2 points ⇒ 2 / 4 points left]

    We are going to start with F^1 and find other angles that are equal to this angle. We will use only the given information.

    F^1=B^ (tan chord theorem)


    STEP: Use the fact that TABC
    [−2 points ⇒ 0 / 4 points left]

    Also, B^=T^3 (corresp s; TABC)

    F^1=T^3
    TIP: Your statements and reasons must work together to prove that F^1=T^3. It is not enough to just choose options that are true.

    Submit your answer as: andandand
  2. Prove that TFAC is a cyclic quadrilateral.

    Here is the diagram again:

    Answer:
    Statement Reason
    FA=CA
    F1=C3
    TFAC is a cyclic quad
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    There are three theorems that we use to prove that a quadrilateral is cyclic:

    1. If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).
    2. If the exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is cyclic (converse ext of cyclic quad).
    3. If a line segment subtends equal angles at two points on the same side of the line segment, then the four points are concyclic (i.e. they form a cyclic quad) (converse s in the same seg).

    STEP: Plan your approach
    [−1 point ⇒ 3 / 4 points left]

    There are three theorems that we use to prove that a quadrilateral is cyclic:

    1. If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).
    2. If the exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is cyclic (converse ext of cyclic quad).
    3. If a line segment subtends equal angles at two points on the same side of the line segment, then the four points are concyclic (i.e. they form a cyclic quad) (converse s in the same seg).

    We have the right information to use the last theorem.


    STEP: Use tans from same pt and s opp equal sides
    [−2 points ⇒ 1 / 4 points left]

    FA=CA (tans from same pt)

    F^1=C^3 (s opp equal sides)

    But, F^1=T^3 (proven above)

    C^3=T^3


    STEP: Use converse s in the same seg
    [−1 point ⇒ 0 / 4 points left]

    TFAC is a cyclic quad (converse s in the same seg)


    Submit your answer as: andandand
  3. Prove that AT bisects FT^C.

    Here is the diagram again:

    Answer:
    Statement Reason
    T2=
    But, T3=
    T2=T3
    Therefore, AT bisects FTC.
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    "Bisect" means to cut into two equal pieces.


    STEP: Use what you have proven already
    [−2 points ⇒ 0 / 2 points left]
    "Bisect" means to cut into two equal pieces.

    We need to prove that T^2=T^3.

    TIP: We have proven that TFAC is a cyclic quad. So we can use s in the same seg.

    T^2=F^1 (s in the same seg)

    But, F^1=T^3 (proven above)

    T^2=T^3

    Therefore, AT bisects FT^C.


    Submit your answer as: andandand

ID is: 4358 Seed is: 812

Circle geometry involving a right angle and trig ratio

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle which goes through B and C. A is a point on chord CB such that OABC. CD and BD are tangents to the circle at C and B respectively.

CB^D=DO^C= 58.4°, and CD=17 units.

Calculate the length of the radius of the circle and give a reason for your answer.

INSTRUCTION: Round your answer to two decimal places.
Answer: The radius of the circle =
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Identify a right-angled triangle that contains a given angle, the given side, and the radius of the circle. Use a trigonometric ratio to find the missing side.


STEP: Identify a right-angled triangle to work in
[−1 point ⇒ 3 / 4 points left]
TIP:

There are several right-angled triangles in this diagram. Right-angled triangles are easy to work with, because we can use trigonometric ratios to find missing sides and angles.

So, we want to identify a right-angled triangle that contains a given angle, the given side, and the side we want to find.

OC is a radius, and CD is a tangent.

OCCD (tan  radius)

This means that ΔOCD is a right-angled triangle which contains:

  • DO^C (a given angle)
  • CD (the given side)
  • OC (the side we want to find)
NOTE: We can also work in ΔOAC and ΔACD, but that would take more steps! This solution shows the shortest way to solve the problem.

STEP: Use a trigonometric ratio to find OC
[−3 points ⇒ 0 / 4 points left]
TIP: In ΔOCD, relative to the given angle DO^C, we were given the opposite side, and want to find the adjacent side.

So, we will work with the ratio:

tanθ=oppositeadjacenttan(58.4°)=17OCOC=17tan(58.4°)OC=10.45846...10.46 units

Submit your answer as: and

ID is: 4358 Seed is: 1325

Circle geometry involving a right angle and trig ratio

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle which goes through Q and R. P is a point on chord RQ such that OPQR. RS and QS are tangents to the circle at R and Q respectively.

RQ^S=SO^R= 67.4°, and RS=11 units.

Calculate the length of the radius of the circle and give a reason for your answer.

INSTRUCTION: Round your answer to two decimal places.
Answer: The radius of the circle =
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Identify a right-angled triangle that contains a given angle, the given side, and the radius of the circle. Use a trigonometric ratio to find the missing side.


STEP: Identify a right-angled triangle to work in
[−1 point ⇒ 3 / 4 points left]
TIP:

There are several right-angled triangles in this diagram. Right-angled triangles are easy to work with, because we can use trigonometric ratios to find missing sides and angles.

So, we want to identify a right-angled triangle that contains a given angle, the given side, and the side we want to find.

OR is a radius, and RS is a tangent.

ORRS (tan  radius)

This means that ΔORS is a right-angled triangle which contains:

  • SO^R (a given angle)
  • RS (the given side)
  • OR (the side we want to find)
NOTE: We can also work in ΔOPR and ΔPRS, but that would take more steps! This solution shows the shortest way to solve the problem.

STEP: Use a trigonometric ratio to find OR
[−3 points ⇒ 0 / 4 points left]
TIP: In ΔORS, relative to the given angle SO^R, we were given the opposite side, and want to find the adjacent side.

So, we will work with the ratio:

tanθ=oppositeadjacenttan(67.4°)=11OROR=11tan(67.4°)OR=4.57885...4.58 units

Submit your answer as: and

ID is: 4358 Seed is: 6807

Circle geometry involving a right angle and trig ratio

Adapted from DBE Nov 2016, Grade 11, P2, Q11
Maths formulas

O is the centre of the circle which goes through Q and R. P is a point on chord RQ such that OPQR. RS and QS are tangents to the circle at R and Q respectively.

RQ^S=SO^R= 63.9°, and the radius of the circle is 5.88 units.

Calculate the length of RS and give a reason for your answer.

INSTRUCTION: Round your answer to two decimal places.
Answer: RS =
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Identify a right-angled triangle that contains a given angle, the given side, and RS. Use a trigonometric ratio to find the missing side.


STEP: Identify a right-angled triangle to work in
[−1 point ⇒ 3 / 4 points left]
TIP:

There are several right-angled triangles in this diagram. Right-angled triangles are easy to work with, because we can use trigonometric ratios to find missing sides and angles.

So, we want to identify a right-angled triangle that contains a given angle, the given side, and the side we want to find.

OR is a radius, and RS is a tangent.

ORRS (tan  radius)

This means that ΔORS is a right-angled triangle which contains:

  • SO^R (a given angle)
  • OR (the given side)
  • RS (the side we want to find)
NOTE: We can also work in ΔOPR and ΔPRS, but that would take more steps! This solution shows the shortest way to solve the problem.

STEP: Use a trigonometric ratio to find RS
[−3 points ⇒ 0 / 4 points left]
TIP: In ΔORS, relative to the given angle SO^R, we were given the adjacent side, and want to find the opposite side.

So, we will work with the ratio:

tanθ=oppositeadjacenttan(63.9°)=RS5.88tan(63.9°)×5.88=RSRS=12.00257...12.00 units

Submit your answer as: and

ID is: 4345 Seed is: 3323

Euclidean geometry: statements of theorems

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

Complete the statement so that it is TRUE:

Two tangents drawn to a circle from the same point outside the circle...
Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which theorem is this question asking you to remember?


STEP: Recall the theorem
[−2 points ⇒ 0 / 2 points left]

The complete theorem statement is:

Two tangents drawn to a circle from the same point outside the circle are equal in length.

Submit your answer as:

ID is: 4345 Seed is: 6704

Euclidean geometry: statements of theorems

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

Complete the statement so that it is TRUE:

The exterior angle of a cyclic quadrilateral is...
Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which theorem is this question asking you to remember?


STEP: Recall the theorem
[−2 points ⇒ 0 / 2 points left]

The complete theorem statement is:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Submit your answer as:

ID is: 4345 Seed is: 1144

Euclidean geometry: statements of theorems

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

Complete the statement so that it is TRUE:

The opposite angles of a cyclic quadrilateral are...
Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which theorem is this question asking you to remember?


STEP: Recall the theorem
[−2 points ⇒ 0 / 2 points left]

The complete theorem statement is:

The opposite angles of a cyclic quadrilateral are supplementary.

Submit your answer as:

ID is: 4413 Seed is: 169

Proving cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, BOC is a diameter of the circle with centre O. ADP is a tangent to the circle at A.

Also, ODAC.

Prove that AOCD is a cyclic quadrilateral.

INSTRUCTION: Make sure your proof as a whole makes sense. True statements that do not help to prove that AOCD is a cyclic quadrilateral will be marked wrong!
Answer:

Let O1=x.

Statement Reason
C1 =90°x (sum of s in Δ)
=90°x
OAD=90°
=x
AOCD is a cyclic quadrilateral
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

STEP: Plan how to prove that the quadrilateral is cyclic
[−1 point ⇒ 4 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

In this question, if we can prove that the angles in the same segment are equal by using other theorems, we know that AOCD has to be cyclic.


STEP: Prove that AOCD is cyclic
[−4 points ⇒ 0 / 5 points left]

Let O^1=x

C^1=90°x (sum of s in Δ)

But, AO=OC (radii)

A^2=90°x (s opp equal sides)

OA^D=90°(tan  radius)

A^1=x

A^1=O^1

AOCD is a cyclic quadrilateral (line subtends equal s) or (converse s in the same seg)

TIP: Both of these reasons would be marked correct, but you only need to give one of them.

Submit your answer as: andandandandand

ID is: 4413 Seed is: 6012

Proving cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, KOL is a diameter of the circle with centre O. JMP is a tangent to the circle at J.

Also, OMJL.

Prove that JOLM is a cyclic quadrilateral.

INSTRUCTION: Make sure your proof as a whole makes sense. True statements that do not help to prove that JOLM is a cyclic quadrilateral will be marked wrong!
Answer:

Let O1=x.

Statement Reason
=90°x
J2 =90°x (s opp equal sides)
OJM=90°
=x
JOLM is a cyclic quadrilateral
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

STEP: Plan how to prove that the quadrilateral is cyclic
[−1 point ⇒ 4 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

In this question, if we can prove that the angles in the same segment are equal by using other theorems, we know that JOLM has to be cyclic.


STEP: Prove that JOLM is cyclic
[−4 points ⇒ 0 / 5 points left]

Let O^1=x

L^1=90°x (sum of s in Δ)

But, JO=OL (radii)

J^2=90°x (s opp equal sides)

OJ^M=90°(tan  radius)

J^1=x

J^1=O^1

JOLM is a cyclic quadrilateral (line subtends equal s) or (converse s in the same seg)

TIP: Both of these reasons would be marked correct, but you only need to give one of them.

Submit your answer as: andandandandand

ID is: 4413 Seed is: 6846

Proving cyclic quads

Adapted from DBE Nov 2016, Grade 11, P2, Q12
Maths formulas

In the diagram below, KOL is a diameter of the circle with centre O. JMP is a tangent to the circle at J.

Also, OMKJ and OMJL.

Prove that JOLM is a cyclic quadrilateral.

INSTRUCTION: Make sure your proof as a whole makes sense. True statements that do not help to prove that JOLM is a cyclic quadrilateral will be marked wrong!
Answer:

Let J1=x.

Statement Reason
=x
=x
JOLM is a cyclic quadrilateral
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

STEP: Plan how to prove that the quadrilateral is cyclic
[−1 point ⇒ 4 / 5 points left]
TIP:

There are three ways to prove that a quadrilateral is cyclic:

  • If a line segment subtends equal angles at two points on the same side of the line segment, then the four points form a cyclic quad (line subtends equal s) or (converse s in the same seg).
  • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic (converse ext of cyclic quad).
  • If the opposite interior angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (converse opp s of cyclic quad).

In this question, if we can prove that the angles in the same segment are equal by using other theorems, we know that JOLM has to be cyclic.


STEP: Prove that JOLM is cyclic
[−4 points ⇒ 0 / 5 points left]

Let J^1=x

K^=x (tan chord theorem)

And, O^1=x (corresp s; OMKJ)

J^1=O^1

JOLM is a cyclic quadrilateral (line subtends equal s) or (converse s in the same seg)

TIP: Both of these reasons would be marked correct, but you only need to give one of them.

Submit your answer as: andandandandand

ID is: 4346 Seed is: 3871

Angle at centre and equal chords

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

O is the centre of the circle which passes through ABCDE.

DO^E=82° and ED=CB.

Calculate the size of:

  1. EA^D
  2. BD^C
Answer: 1. EA^D= °
2. BD^C= °
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Finding EA^D will help you to find BD^C . Which Euclidean geometry theorem can you apply to connect EA^D to a known angle?


STEP: Find EA^D
[−2 points ⇒ 2 / 4 points left]

We will focus on EA^D first, and try to connect it to our known angle, DO^E.

EA^D is an angle at the circumference. It was subtended by the same chord as DO^E.

Therefore, DO^E is double EA^D.

EA^D=41° ( at centre =2× at circumference)


STEP: Find BD^C
[−2 points ⇒ 0 / 4 points left]

Now that we know EA^D=41° we will try to connect it to BD^C .

BD^C and EA^D are both angles at the circumference. They were subtended by equal chords.

Therefore, BD^C is equal to EA^D.

BD^C=41° (equal chords; equal s)


Submit your answer as: andandand

ID is: 4346 Seed is: 8773

Angle at centre and equal chords

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

O is the centre of the circle which passes through ABCDE.

DO^E=100° and ED=CB.

Calculate the size of:

  1. EA^D
  2. BD^C
Answer: 1. EA^D= °
2. BD^C= °
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Finding EA^D will help you to find BD^C . Which Euclidean geometry theorem can you apply to connect EA^D to a known angle?


STEP: Find EA^D
[−2 points ⇒ 2 / 4 points left]

We will focus on EA^D first, and try to connect it to our known angle, DO^E.

EA^D is an angle at the circumference. It was subtended by the same chord as DO^E.

Therefore, DO^E is double EA^D.

EA^D=50° ( at centre =2× at circumference)


STEP: Find BD^C
[−2 points ⇒ 0 / 4 points left]

Now that we know EA^D=50° we will try to connect it to BD^C .

BD^C and EA^D are both angles at the circumference. They were subtended by equal chords.

Therefore, BD^C is equal to EA^D.

BD^C=50° (equal chords; equal s)


Submit your answer as: andandand

ID is: 4346 Seed is: 5604

Angle at centre and equal chords

Adapted from DBE Nov 2016 Grade 11, P2, Q9
Maths formulas

O is the centre of the circle which passes through ABDE.

BO^D=74° and ED=DB.

Calculate the size of:

  1. DO^E
  2. EA^D
Answer: 1. DO^E= °
2. EA^D= °
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Finding DO^E will help you to find EA^D . Which Euclidean geometry theorem can you apply to connect DO^E to a known angle?


STEP: Find DO^E
[−2 points ⇒ 2 / 4 points left]

We will focus on DO^E first, and try to connect it to our known angle, BO^D.

DO^E and BO^D are both angles at the centre. They were subtended by equal chords.

Therefore, DO^E is equal to BO^D.

DO^E=74° (equal chords; equal s)


STEP: Find EA^D
[−2 points ⇒ 0 / 4 points left]

Now that we know DO^E=74° we will try to connect it to EA^D .

EA^D is an angle at the circumference. It was subtended by the same chord as DO^E.

Therefore, DO^E is double EA^D.

EA^D=37° ( at centre =2× at circumference)


Submit your answer as: andandand

ID is: 4377 Seed is: 7031

Proving that a point is the centre of a circle

Adapted from DBE Nov 2015, Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices A, F and C of ΔAFC are concyclic. BF and BC are tangents to the circle at F and C respectively. T is a point on AF such that TBAC. FC cuts TB at E.

It is given that TFBC is a cyclic quadrilateral.

  1. If it is given that F^1+B^1=90°, prove that TC^B=TF^B=90°.

    INSTRUCTION: Complete the proof as it is set out here. Read through the whole proof before you start.
    Answer:
    Statement Reason
    F2=B1
    =90°
    =90°
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You are given that F^1+B^1=90°. From the diagram, TF^B=F^1+F^2.


    STEP: Use s in the same seg
    [−2 points ⇒ 2 / 4 points left]

    We are given that F^1+B^1=90°. From the diagram, TF^B=F^1+F^2.

    We need to prove that B^1=F^2:

    F^2=B^1 (s in the same seg)


    STEP: Use the information you have been given
    [−1 point ⇒ 1 / 4 points left]

    But, F^1+B^1=90° (given)

    F^1+F^2=90°

    TF^B=90°


    STEP: Use opp s of cyclic quad
    [−1 point ⇒ 0 / 4 points left]

    TC^B+TF^B=180° (opp s of cyclic quad)

    TC^B=90°

    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand
  2. Hence, prove that T is the centre of the circle through A, F and C.

    Here is the question diagram again:

    Answer:

    AF is a diameter of the circle .
    But, the line , produced to the circumference, is a diameter of the circle .

    Therefore:

    T is the centre of the circle.

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You have already proven that TC^B=TF^B=90°. To prove that T is the centre, you must prove that T is the point of intersection of two diameters.


    STEP: Prove that T is the point of intersection of two diameters
    [−3 points ⇒ 0 / 3 points left]

    We have already proven that TC^B=TF^B=90°. To prove that T is the centre, we will prove that T is the point of intersection of two diameters.

    NOTE: It is not enough to only prove that AF is a diameter. There are infinitely many points on a diameter that are not the centre of a circle. But, diameters only cross each other at the centre of the circle. So a point where two diameters intersect must be the centre of the circle.

    AF is a diameter of the circle (converse tan diameter).

    But, the line TC, produced to the circumference, is also a diameter of the circle (converse tan diameter).

    T is the point where two diameters meet.

    Therefore, T is the centre of the circle.

    Use the correct theorem!

    • We use converse tan diameter to prove a line is a tangent or a diameter (because we know the 90° angle).
    • We use tan diameter to prove that the angle is equal to 90° (because we know the tangent and the diameter).
    A ✘ The circle has two diameters.
    Every circle has infinitely many diameters. This sentence does not add anything to the proof.
    B ✘ TCCB and TFFB (tan radius)
    We have already proven above that TCCB and AFFB; this option just repeats that fact (with the wrong reason). We still need to explain why this means that T is the centre.
    C ✔ T is the point where two diameters meet.
    The point where two diameters intersect is the centre of the circle.
    D ✘ The diameters are perpendicular to the tangents (converse tan diameter).
    We have already proven above that AF is a diameter and TC lies on a diameter; this option just repeats that fact. We still need to explain why this means that T is the centre.

    Submit your answer as: andandand

ID is: 4377 Seed is: 6979

Proving that a point is the centre of a circle

Adapted from DBE Nov 2015, Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices C, E and A of ΔCEA are concyclic. FE and FA are tangents to the circle at E and A respectively. T is a point on CE such that TFCA. EA cuts TF at B.

It is given that TEFA is a cyclic quadrilateral. Also, T^2=T^3=C^.

  1. If it is given that TE is a tangent to the circle through E, B and F, prove that TE=TA.

    Answer:
    Statement Reason
    E2=F2
    Also, F2=A2
    TE=TA
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    TE and TA are sides in ΔTEA. Use the fact that TE is a tangent to prove that the base angles are equal.


    STEP: Use tan chord theorem
    [−2 points ⇒ 2 / 4 points left]
    TIP: The question tells you that TE is a tangent to the circle through E, B and F. You should draw in the circle on your question paper.
    We are going to use the fact that TE is a tangent to prove that E^2=A^2 (and hence that TE=TA).

    E^2=F^2 (tan chord theorem)


    STEP: Use s in the same seg
    [−1 point ⇒ 1 / 4 points left]

    Also, F^2=A^2 (s in the same seg)


    STEP: Use sides opp equal s
    [−1 point ⇒ 0 / 4 points left]

    E^2=A^2

    TE=TA (sides opp equal s)

    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand
  2. Hence, prove that T is the centre of the circle through C, E and A.

    Here is the question diagram again:

    Answer:

    A1=T2

    (sides opp equal s)

    Therefore:

    T is the centre of the circle.

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You have already proven that TE=TA. To prove that T is the centre, you also need to prove that TC=TA.


    STEP: Prove that TC=TA
    [−3 points ⇒ 0 / 3 points left]

    We have already proven that TE=TA. To prove that T is the centre, we also need to prove that TC=TA.

    TIP: It is given that T^2=T^3=C^. So if we can prove that A^1 is equal to either T^2 or T^3, then we will know that A^1=C^.

    A^1=T^2 (alt s; CATF)

    But, T^2=C^ (given)

    A^1=C^

    TC=TA (sides opp equal s)

    TC=TA=TE

    T is a point the same distance from C, E and A on the circle.

    T is the centre of the circle.

    A ✘ T is a point the same distance from E and A on the circle.

    We can't be sure T is the centre if we only know TA=TE. There are infinitely many points in a circle that are the same distance from two points on the circumference. In the diagram below, D is the same distance from A and E, but it is not the centre of the circle.

    However, if a point is the same distance from at least three points on the circumference, then that point is the centre of the circle.

    B ✔ TC=TA=TE
    This means that T is a point the same distance from C, E and A on the circle. So, T is the same distance from three points on the circumference.
    C ✘ TAAF and TEEF (tan radius)
    This option uses the fact that T is the centre of the circle. But, this is what we are trying to prove.
    D ✘ TC and TA are radii (converse tan radius)
    The reason converse tan radius can be used to prove that a line is part of a radius. To do so, we need to prove that the line is perpendicular to a tangent. But we have not done that in this proof.

    Submit your answer as: andandand

ID is: 4377 Seed is: 8854

Proving that a point is the centre of a circle

Adapted from DBE Nov 2015, Grade 11, P2, Q11
Maths formulas

In the diagram, the vertices C, B and E of ΔCBE are concyclic. FB and FE are tangents to the circle at B and E respectively. T is a point on CB such that TFCE. BE cuts TF at A.

It is given that TBFE is a cyclic quadrilateral.

  1. If it is given that E^2+T^3=90°, prove that TE^F=TB^F=90°.

    INSTRUCTION: Complete the proof as it is set out here. Read through the whole proof before you start.
    Answer:
    Statement Reason
    E3=T3
    =90°
    =90°
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You are given that E^2+T^3=90°. From the diagram, TE^F=E^2+E^3.


    STEP: Use s in the same seg
    [−2 points ⇒ 2 / 4 points left]

    We are given that E^2+T^3=90°. From the diagram, TE^F=E^2+E^3.

    We need to prove that T^3=E^3:

    E^3=T^3 (s in the same seg)


    STEP: Use the information you have been given
    [−1 point ⇒ 1 / 4 points left]

    But, E^2+T^3=90° (given)

    E^2+E^3=90°

    TE^F=90°


    STEP: Use opp s of cyclic quad
    [−1 point ⇒ 0 / 4 points left]

    TE^F+TB^F=180° (opp s of cyclic quad)

    TB^F=90°

    TIP: When choosing the correct options for this proof, it is not enough to select any true geometry fact. You must choose the options that combine to form the correct proof.

    Submit your answer as: andandand
  2. Hence, prove that T is the centre of the circle through C, B and E.

    Here is the question diagram again:

    Answer:

    CB is a diameter of the circle .
    But, the line ET, produced to the circumference, is a of the circle

    Therefore:

    T is the centre of the circle.

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You have already proven that TE^F=TB^F=90°. To prove that T is the centre, you must prove that T is the point of intersection of two diameters.


    STEP: Prove that T is the point of intersection of two diameters
    [−3 points ⇒ 0 / 3 points left]

    We have already proven that TE^F=TB^F=90°. To prove that T is the centre, we will prove that T is the point of intersection of two diameters.

    NOTE: It is not enough to only prove that CB is a diameter. There are infinitely many points on a diameter that are not the centre of a circle. But, diameters only cross each other at the centre of the circle. So a point where two diameters intersect must be the centre of the circle.

    CB is a diameter of the circle (converse tan diameter).

    But, the line TE, produced to the circumference, is also a diameter of the circle (converse tan diameter).

    Two diameters of the circle intersect at T.

    Therefore, T is the centre of the circle.

    Use the correct theorem!

    • We use converse tan diameter to prove a line is a tangent or a diameter (because we know the 90° angle).
    • We use tan diameter to prove that the angle is equal to 90° (because we know the tangent and the diameter).
    A ✘ The circle has two diameters.
    Every circle has infinitely many diameters. This sentence does not add anything to the proof.
    B ✔ Two diameters of the circle intersect at T.
    The point where two diameters intersect is the centre of the circle.
    C ✘ TEEF and TBBF (tan radius)
    We have already proven above that TEEF and CBBF; this option just repeats that fact (with the wrong reason). We still need to explain why this means that T is the centre.
    D ✘ The diameters are perpendicular to the tangents (converse tan diameter).
    We have already proven above that CB is a diameter and TE lies on a diameter; this option just repeats that fact. We still need to explain why this means that T is the centre.

    Submit your answer as: andandand